The sphere starts from rest at `\theta = 0 ^\circ` and rotates with an angular acceleration of `\alpha = \left(4\theta + 1\right)\text{ rad/s}^2`, where `\theta` is in radians.

Determine the magnitudes of the velocity and acceleration of point `P` on the sphere at the instant `\theta = 6\text{ rad}`.

Angular velocity at `\theta = 6\text{ rad}`.

\begin{aligned} \alpha &= 4\theta + 1 \\[1pt] \omega\frac{d\omega}{d\theta} &= 4\theta + 1 \\[1pt] \int_0^{\omega} \omega\ d\omega &= \int_0^{6} \left(4\theta + 1\right)\ d\theta \\[1pt] \left[\frac{\omega^2}{2}\right]_0^{\omega} &= \Big[2\theta^2 + \theta\Big]_0^{6} \\[1pt] \frac{\omega^2}{2} &= 78 \\[1pt] \omega &= 12.49\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity at `P`.

\begin{aligned} v_P &= r_P\thinspace\omega \\[1pt] v_P &= \left(0.67\cos 30 ^\circ \right) 12.49 \color{green}\text{ // }r = 8\text{ in} = 0.67\text{ ft} \\[1pt] v_P &= 7.21\text{ ft/s} \\[1pt] \end{aligned}

Angular acceleration at `\theta = 6\text{ rad}`.

\begin{aligned} \alpha &= 4\theta + 1 \\[1pt] \alpha &= \left(4 \times 6\right) + 1 \\[1pt] \alpha &= 25\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear acceleration at `P`.

\begin{aligned} a_P &= \left(a_P\right)_t + \left(a_P\right)_c \\[1pt] a_P &= r_P \thinspace\alpha + r_P\thinspace \omega^2 \\[1pt] a_P &= \left(0.67\cos 30 ^\circ \right) \left(25\right) + \left(0.67\cos 30 ^\circ \right) \left(12.49\right)^2 \color{green}\text{ // }r = 8\text{ in} = 0.67\text{ ft} \\[1pt] a_P &= 14.43 + 90.07 \\[1pt] a_P &= 91.22\text{ ft/s}^2 \\[1pt] \end{aligned}

The rod assembly is supported by ball-and-socket joints at `A` and `B`.

At the instant shown it is rotating about the `y` axis with an angular velocity `\omega = 5\text{ rad/s}` and has an angular acceleration `\alpha = 8\text{ rad/s}^2`.

Determine the magnitudes of the velocity and acceleration of point `C` at this instant.

Solve the problem using Cartesian vectors and Eqs. 16-9 and 16-13.

Velocity.

\begin{aligned} \vec v_C &= \omega \times \vec r_{AC} \\[1pt] \vec v_C &= \begin{bmatrix} 0 \\ 5 \\ 0 \\ \end{bmatrix} \times \begin{bmatrix} -0.4 \\ 0 \\ 0.3 \\ \end{bmatrix} \\[1pt] \hline \vec v_C &= \begin{bmatrix} 1.5 \\ 0 \\ 2 \\ \end{bmatrix} \text{ m/s} \\[1pt] v_C &= \sqrt{\left(1.5\right)^2 + \left(0\right)^2 + \left(2\right)^2} = 2.5\text{ m/s} \\[1pt] \end{aligned}

Acceleration.

\begin{aligned} \vec a_C &= \left(\vec a_C\right)_t + \left(\vec a_C\right)_c \\[1pt] \vec a_C &= \left(\alpha \times \vec r_{AC}\right) + \left(\vec\omega \times \vec v_C\right) \\[1pt] \vec a_C &= \left( \begin{bmatrix} 0 \\ 8 \\ 0 \\ \end{bmatrix} \times \begin{bmatrix} -0.4 \\ 0 \\ 0.3 \\ \end{bmatrix} \right) + \left( \begin{bmatrix} 0 \\ 5 \\ 0 \\ \end{bmatrix} \times \begin{bmatrix} 1.5 \\ 0 \\ 2 \\ \end{bmatrix} \right) \\[1pt] \vec a_C &= \begin{bmatrix} 2.4 \\ 0 \\ 3.2 \\ \end{bmatrix} + \begin{bmatrix} 10 \\ 0 \\ -7.5 \\ \end{bmatrix} \\[1pt] \hline \vec a_C &= \begin{bmatrix} 12.4 \\ 0 \\ -4.3 \\ \end{bmatrix} \text{ m/s}^2 \\[1pt] a_C &= \sqrt{\left(12.4\right)^2 + \left(0\right)^2 + \left(-4.3\right)^2} = 13.12\text{ m/s}^2 \\[1pt] \end{aligned}

At the instant shown, the shaft and plate rotates with an angular velocity of `\omega = 14\text{ rad/s}` and angular acceleration of `\alpha = 7\text{ rad/s}^2`.

Determine the velocity and acceleration of point `D` located on the corner of the plate at this instant.

Express the result in Cartesian vector form.

Angular velocity vector `\vec\omega`.

\begin{aligned} \vec\omega &= \omega\thinspace \hat r_{OA} \\[1pt] \vec\omega &= 14 \left(\frac{\vec r_{OA}}{\parallel \vec r_{OA} \parallel}\right) \\[1pt] \vec\omega &= 14 \left( \begin{bmatrix} -0.3 \\ 0.2 \\ 0.6 \\ \end{bmatrix} \frac{1}{0.7}\right) \\[1pt] \vec\omega &= \begin{bmatrix} -6 \\ 4 \\ 12 \\ \end{bmatrix} \text{ rad/s} \\[1pt] \end{aligned}

Linear velocity `\vec v_D`.

\begin{aligned} \vec v_D &= \vec\omega \times \vec r_D \\[1pt] \vec v_D &= \begin{bmatrix} -6 \\ 4 \\ 12 \\ \end{bmatrix} \times \begin{bmatrix} 0.3 \\ -0.4 \\ 0 \\ \end{bmatrix} \\[1pt] \vec v_D &= \begin{bmatrix} 4.8 \\ 3.6 \\ 1.2 \\ \end{bmatrix} \text{ m/s} \\[1pt] \end{aligned}

Angular acceleration vector `\vec\alpha`.

\begin{aligned} \vec\alpha &= \alpha\thinspace \hat r_{OA} \\[1pt] \vec\alpha &= 7 \left(\frac{\vec r_{OA}}{\parallel \vec r_{OA} \parallel}\right) \\[1pt] \vec\alpha &= 7 \left( \begin{bmatrix} -0.3 \\ 0.2 \\ 0.6 \\ \end{bmatrix} \frac{1}{0.7}\right) \\[1pt] \vec\alpha &= \begin{bmatrix} -3 \\ 2 \\ 6 \\ \end{bmatrix} \text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration `\left(\vec a_D\right)_t`.

\begin{aligned} \left(\vec a_D\right)_t &= \vec\alpha \times \vec r_D \\[1pt] \left(\vec a_D\right)_t &= \begin{bmatrix} -3 \\ 2 \\ 6 \\ \end{bmatrix} \times \begin{bmatrix} 0.3 \\ -0.4 \\ 0 \\ \end{bmatrix} \\[1pt] \left(\vec a_D\right)_t &= \begin{bmatrix} 2.4 \\ 1.8 \\ 0.6 \\ \end{bmatrix} \text{ m/s} \\[1pt] \end{aligned}

Centripetal acceleration `\left(\vec a_D\right)_c`.

\begin{aligned} \left(\vec a_D\right)_c &= \vec\omega \times \vec v_D \\[1pt] \left(\vec a_D\right)_c &= \begin{bmatrix} -6 \\ 4 \\ 12 \\ \end{bmatrix} \times \begin{bmatrix} 4.8 \\ 3.6 \\ 1.2 \\ \end{bmatrix} \\[1pt] \left(\vec a_D\right)_c &= \begin{bmatrix} -38.4 \\ 64.8 \\ -40.8 \\ \end{bmatrix} \text{ m/s}^2 \\[1pt] \end{aligned}

Net acceleration `\vec a_D`.

\begin{aligned} \vec a_D &= \left(\vec a_D\right)_t + \left(\vec a_D\right)_c \\[1pt] \vec a_D &= \begin{bmatrix} 2.4 \\ 1.8 \\ 0.6 \\ \end{bmatrix} + \begin{bmatrix} -38.4 \\ 64.8 \\ -40.8 \\ \end{bmatrix} \\[1pt] \vec a_D &= \begin{bmatrix} -36 \\ 66.6 \\ -40.2 \\ \end{bmatrix} \text{ m/s}^2 \\[1pt] \end{aligned}

If the shaft and plate rotates with a constant angular velocity of `v = 14\text{ rad/s}`, determine the velocity and acceleration of point `C` located on the corner of the plate at the instant shown.

Express the result in Cartesian vector form.

Angular velocity vector `\vec\omega`.

\begin{aligned} \vec\omega &= \omega\thinspace \hat r_{OA} \\[1pt] \vec\omega &= 14 \left(\frac{\vec r_{OA}}{\parallel \vec r_{OA} \parallel}\right) \\[1pt] \vec\omega &= 14 \left( \begin{bmatrix} -0.3 \\ 0.2 \\ 0.6 \\ \end{bmatrix} \frac{1}{0.7}\right) \\[1pt] \vec\omega &= \begin{bmatrix} -6 \\ 4 \\ 12 \\ \end{bmatrix} \text{ rad/s} \\[1pt] \end{aligned}

Linear velocity `\vec v_C`.

\begin{aligned} \vec v_C &= \vec\omega \times \vec r_C \\[1pt] \vec v_C &= \begin{bmatrix} -6 \\ 4 \\ 12 \\ \end{bmatrix} \times \begin{bmatrix} -0.3 \\ 0.4 \\ 0 \\ \end{bmatrix} \\[1pt] \vec v_C &= \begin{bmatrix} -4.8 \\ -3.6 \\ -1.2 \\ \end{bmatrix} \text{ m/s} \\[1pt] \end{aligned}

Centripetal acceleration `\left(\vec a_C\right)_c`.

\begin{aligned} \left(\vec a_C\right)_c &= \vec\omega \times \vec v_C \\[1pt] \left(\vec a_C\right)_c &= \begin{bmatrix} -6 \\ 4 \\ 12 \\ \end{bmatrix} \times \begin{bmatrix} -4.8 \\ -3.6 \\ -1.2 \\ \end{bmatrix} \\[1pt] \left(\vec a_C\right)_c &= \begin{bmatrix} 38.4 \\ -64.8 \\ 40.8 \\ \end{bmatrix} \text{ m/s}^2 \\[1pt] \end{aligned}

For the outboard motor in Prob. 16-24, determine the magnitude of the velocity and acceleration of point `P` located on the tip of the propeller at the instant `t = 0.75\text{ s}`.

Angular velocity of gear `A` at `t = 0.75\text{ seconds}`.

\begin{aligned} \alpha &= 400t^3 \\[1pt] \frac{d\omega}{dt} &= 400t^3 \\[1pt] \int_0^{\omega_A} d\omega &= \int_0^{0.75} 400t^3\ dt \\[1pt] \Big[\omega\Big]_0^{\omega_A} &= \Big[100t^4\Big]_0^{0.75} \\[1pt] \omega_A &= 31.64\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `A` at `t = 0.75\text{ seconds}`.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 0.5 \times 31.64 \\[1pt] v_A &= 15.82\text{ in/s} \\[1pt] \end{aligned}

Angular velocity of gear `B` at `t = 0.75\text{ seconds}`.

\begin{aligned} v_A &= r_B\omega_B \\[1pt] \omega_B &= \frac{v_A}{r_B} \\[1pt] \omega_B &= \frac{15.82}{1.2} \\[1pt] \omega_B &= 13.18\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of point `P`.

\begin{aligned} v_P &= r_P\omega_B \\[1pt] v_P &= 2.2 \times 13.18 \\[1pt] v_P &= 29\text{ in/s} = 2.42\text{ ft/s} \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} \alpha_A &= 400t^3 \\[1pt] \alpha_A &= 400 \times 0.75^3 \\[1pt] \alpha_A &= 168.75\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear acceleration.

\begin{aligned} a_A &= r_A\alpha_A \\[1pt] a_A &= 0.5 \times 168.75 \\[1pt] a_A &= 84.38\text{ in/s}^2 \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} a_A &= r_B\alpha_B \\[1pt] \alpha_B &= \frac{a_A}{r_B} \\[1pt] \alpha_B &= \frac{84.38}{1.2} \\[1pt] \alpha_B &= 70.31\text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration.

\begin{aligned} \left(a_P\right)_t &= r_P \alpha_B \\[1pt] \left(a_P\right)_t &= 2.2 \times 70.31 \\[1pt] \left(a_P\right)_t &= 154.69\text{ in/s}^2 \\[1pt] \end{aligned}

Centripetal acceleration.

\begin{aligned} \left(a_P\right)_c &= r_P \left(\omega_B\right)^2 \\[1pt] \left(a_P\right)_c &= 2.2 \left(13.18\right)^2 \\[1pt] \left(a_P\right)_c &= 382.38\text{ in/s}^2 \\[1pt] \end{aligned}

Net acceleration.

\begin{aligned} a_P &= \sqrt{\left(a_P\right)_t^2 + \left(a_P\right)_c^2} \\[1pt] a_P &= \sqrt{154.69^2 + 382.38^2} \\[1pt] a_P &= 412.48\text{ in/s}^2 = 34.37\text{ ft/s}^2 \\[1pt] \end{aligned}

The gear `A` on the drive shaft of the outboard motor has a radius `r_A = 0.5\text{ in}.` and the meshed pinion gear `B` on the propeller shaft has a radius `r_B = 1.2\text{ in}`.

Determine the angular velocity of the propeller in `t = 1.5\text{ s}`, if the drive shaft rotates with an angular acceleration `\alpha = \left(400t^3\right)\text{ rad/s}^2`, where `t` is in seconds.

The propeller is originally at rest and the motor frame does not move.

Angular velocity of gear `A` at `t = 1.5\text{ seconds}`.

\begin{aligned} \alpha &= 400t^3 \\[1pt] \frac{d\omega}{dt} &= 400t^3 \\[1pt] \int_0^{\omega_A} d\omega &= \int_0^{1.5} 400t^3\ dt \\[1pt] \Big[\omega\Big]_0^{\omega_A} &= \Big[100t^4\Big]_0^{1.5} \\[1pt] \omega_A &= 506.25\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `A` at `t = 1.5\text{ seconds}`.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 0.5 \times 506.25 \\[1pt] v_A &= 253.125\text{ in/s} \\[1pt] \end{aligned}

Angular velocity of gear `B` at `t = 1.5\text{ seconds}`.

\begin{aligned} v_A &= r_B\omega_B \\[1pt] \omega_B &= \frac{v_A}{r_B} \\[1pt] \omega_B &= \frac{253.125}{1.2} \\[1pt] \omega_B &= 210.94\text{ rad/s} \\[1pt] \end{aligned}

Angular velocity of the propeller equals the angular velocity gear `B`.

If the motor turns gear `A` with an angular acceleration of `a_A = 3\text{ rad/s}^2` when the angular velocity is `v_A = 60\text{ rad/s}`, determine the angular acceleration and angular velocity of gear `D`.

Linear velocity.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 0.04 \text{ meters} \times 60 \text{ rad/s} \\[1pt] v_A &= 2.4\text{ m/s} \\[1pt] \end{aligned}

Linear acceleration.

\begin{aligned} a_A &= r_A\alpha_A \\[1pt] a_A &= 0.04 \text{ meters} \times 3 \text{ rad/s}^2 \\[1pt] a_A &= 0.12\text{ m/s}^2 \\[1pt] \end{aligned}

Angular velocity.

\begin{aligned} \omega_B &= \frac{v_A}{r_B} \\[1pt] \omega_B &= \frac{2.4}{0.1} \\[1pt] \omega_B &= 24\text{ rad/s} \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} \alpha_B &= \frac{a_A}{r_B} \\[1pt] \alpha_B &= \frac{0.12}{0.1} \\[1pt] \alpha_B &= 1.2\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear velocity.

\begin{aligned} v_C &= r_C\omega_B \\[1pt] v_C &= 0.05 \times 24 \\[1pt] v_C &= 1.2\text{ m/s} \\[1pt] \end{aligned}

Linear acceleration.

\begin{aligned} a_C &= r_C\alpha_B \\[1pt] a_C &= 0.05 \times 1.2 \\[1pt] a_C &= 0.06\text{ m/s}^2 \\[1pt] \end{aligned}

Angular velocity.

\begin{aligned} \omega_D &= \frac{v_C}{r_D} \\[1pt] \omega_D &= \frac{1.2}{0.1} \\[1pt] \omega_D &= 12\text{ rad/s} \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} \alpha_D &= \frac{a_C}{r_D} \\[1pt] \alpha_D &= \frac{0.06}{0.1} \\[1pt] \alpha_D &= 0.6\text{ rad/s} \\[1pt] \end{aligned}

If the motor turns gear `A` with an angular acceleration of `\alpha_A = 2\text{ rad/s}^2` when the angular velocity is `\omega_A = 20\text{ rad/s}`, determine the angular acceleration and angular velocity of gear `D`.

Linear velocity.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 0.04 \text{ meters} \times 20 \text{ rad/s} \\[1pt] v_A &= 0.8\text{ m/s} \\[1pt] \end{aligned}

Linear acceleration.

\begin{aligned} a_A &= r_A\alpha_A \\[1pt] a_A &= 0.04 \text{ meters} \times 2 \text{ rad/s}^2 \\[1pt] a_A &= 0.08\text{ m/s}^2 \\[1pt] \end{aligned}

Angular velocity.

\begin{aligned} \omega_B &= \frac{v_A}{r_B} \\[1pt] \omega_B &= \frac{0.8}{0.1} \\[1pt] \omega_B &= 8\text{ rad/s} \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} \alpha_B &= \frac{a_A}{r_B} \\[1pt] \alpha_B &= \frac{0.08}{0.1} \\[1pt] \alpha_B &= 0.8\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear velocity.

\begin{aligned} v_C &= r_C\omega_B \\[1pt] v_C &= 0.05 \times 8 \\[1pt] v_C &= 0.4\text{ m/s} \\[1pt] \end{aligned}

Linear acceleration.

\begin{aligned} a_C &= r_C\alpha_B \\[1pt] a_C &= 0.05 \times 0.8 \\[1pt] a_C &= 0.04\text{ m/s}^2 \\[1pt] \end{aligned}

Angular velocity.

\begin{aligned} \omega_D &= \frac{v_C}{r_D} \\[1pt] \omega_D &= \frac{0.4}{0.1} \\[1pt] \omega_D &= 4\text{ rad/s} \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} \alpha_D &= \frac{a_C}{r_D} \\[1pt] \alpha_D &= \frac{0.04}{0.1} \\[1pt] \alpha_D &= 0.4\text{ rad/s} \\[1pt] \end{aligned}

The motor turns the disk with an angular velocity of `\omega = \left(5t^2 + 3t\right)\text{ rad/s}`, where `t` is in seconds.

Determine the magnitudes of the velocity and the `n` and `t` components of acceleration of the point `A` on the disk when `t = 3\text{ s}`.

Angular velocity of the disk at `t = 3\text{ seconds}`.

\begin{aligned} \omega &= 5t^2 + 3t \\[1pt] \omega &= 5 \left(3^2\right) + \left(3 \times 3\right) \\[1pt] \omega &= 54\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of the disk at point `A` when `t = 3\text{ seconds}`.

\begin{aligned} v_A &= r_A\omega \\[1pt] v_A &= 150 \times 54 \\[1pt] v_A &= 8100\text{ mm/s} \\[1pt] \end{aligned}

Angular acceleration of the disk at `t = 3\text{ seconds}`.

\begin{aligned} \omega &= 5t^2 + 3t \\[1pt] \frac{d\omega}{dt} &= \frac{d}{dt}\Big[5t^2 + 3t\Big] \\[1pt] \alpha &= 10t + 3 \\[1pt] \alpha &= \left(10 \times 3\right) + 3 \\[1pt] \alpha &= 33\text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration.

\begin{aligned} a_t &= r_A \alpha \\[1pt] a_t &= 150 \times 33 \\[1pt] a_t &= 4950\text{ mm/s}^2 \\[1pt] \end{aligned}

Normal acceleration.

\begin{aligned} a_n &= r_A\omega^2 \\[1pt] a_n &= 150 \times 54^2 \\[1pt] a_n &= 437400\text{ mm/s}^2 \\[1pt] \end{aligned}

A motor gives gear `A` an angular acceleration of `\alpha_A = \left(4t^3\right)\text{ rad/s}^2`, where `t` is in seconds.

If this gear is initially turning at `\left(\omega_A\right)_0 = 20\text{ rad/s}`, determine the angular velocity of gear `B` when `t = 2\text{ s}`.

Angular velocity of gear `A`.

\begin{aligned} \alpha_A &= 4t^3 \\[1pt] \frac{d\omega_A}{dt} &= 4t^3 \\[1pt] \int_{20}^{\omega_A} d\omega_A &= \int_0^{2} 4t^3 \ dt \\[1pt] \omega_A - 20 &= 2^4 \\[1pt] \omega_A &= 36\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `A`.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 0.05 \times 36 \\[1pt] v_A &= 1.8\text{ m/s} \\[1pt] \end{aligned}

Angular velocity of gear `B`.

\begin{aligned} \omega_B &= \frac{v_A}{r_B} \\[1pt] \omega_B &= \frac{1.8}{0.15} \\[1pt] \omega_B &= 12\text{ rad/s} \\[1pt] \end{aligned}

The vacuum cleaner’s armature shaft `S` rotates with an angular acceleration of `\alpha = 4\omega^{3/4}\text{ rad/s}^2`, where `\omega` is in `\text{rad/s}`.

Determine the brush’s angular velocity when `t = 4\text{ s}`, starting from `\omega_0 = 1\text{ rad/s}`, at `\theta = 0`.

The radii of the shaft and the brush are `0.25\text{ in.}` and `1\text{ in.}`, respectively.

Neglect the thickness of the drive belt.

Angular velocity of the shaft `S`.

\begin{aligned} \alpha &= 4\omega^{3/4} \\[1pt] \frac{d\omega}{dt} &= 4\omega^{3/4} \\[1pt] \frac{d\omega}{4\omega^{3/4}} &= dt \\[1pt] \frac{1}{4} \int_{1}^{\omega_{\text{shaft}}} \omega^{-3/4}\ d\omega &= \int_0^{4} dt \\[1pt] \frac{1}{4} \left[4\omega^{1/4}\right]_{1}^{\omega_{\text{shaft}}} &= \Big[t\Big]_0^{4} \\[1pt] \left({\omega_{\text{shaft}}}\right)^{1/4} - 1 &= 4 \\[1pt] \hline \omega_{\text{shaft}} &= 625\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of the shaft `S`.

\begin{aligned} v_{\text{shaft}} &= r_{\text{shaft}} \omega_{\text{shaft}}\\[1pt] v_{\text{shaft}} &= 0.25 \times 625 \\[1pt] v_{\text{shaft}} &= 156.25\text{ in/s} \\[1pt] \end{aligned}

Angular velocity of the brush.

\begin{aligned} \omega_{\text{brush}} &= \frac{v_{\text{shaft}}}{r_{\text{brush}}} \\[1pt] \omega_{\text{brush}} &= \frac{156.25}{1} \\[1pt] \omega_{\text{brush}} &= 156.25\text{ rad/s} \\[1pt] \end{aligned}

A motor gives gear `A` an angular acceleration of `\alpha = \left(2t^3\right)\text{ rad/s}^2`, where `t` is in seconds.

If this gear is initially turning at `\omega_A = 15\text{ rad/s}`, determine the angular velocity of gear `B` when `t = 3\text{ s}`.

Angular velocity of gear `A` at `t = 3\text{ seconds}`.

\begin{aligned} \alpha &= 2t^3 \\[1pt] \frac{d\omega}{dt} &= 2t^3 \\[1pt] \int_{15}^{\omega_A} d\omega &= \int_0^3 2t^3\ dt \\[1pt] \omega_A - 15 &= 40.5 \\[1pt] \omega_A &= 55.5\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `A` at `t = 3\text{ seconds}`.

\begin{aligned} v_A &= r_A \omega_A \\[1pt] v_A &= 100 \times 55.5 \\[1pt] v_A &= 5550\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of gear `B` at `t = 3\text{ seconds}`.

\begin{aligned} \omega_A &= \frac{v_A}{r_B} \\[1pt] \omega_A &= \frac{5550}{175} \\[1pt] \omega_A &= 31.71\text{ rad/s} \\[1pt] \end{aligned}

A motor gives gear `A` an angular acceleration of `\alpha_A = \left(2 + 0.006\theta^2\right)\text{ rad/s}^2`, where `\theta` is in radians.

If this gear is initially turning at `\omega_A = 15\text{ rad/s}`, determine the angular velocity of gear `B` after `A` undergoes an angular displacement of `10\text{ rev}`.

Angular velocity function of gear `A`.

\begin{aligned} \alpha_A &= 2 + 0.006\theta^2 \\[1pt] \omega_A\frac{d\omega_A}{d\theta} &= 2 + 0.006\theta^2 \\[1pt] \int_{15}^{\omega_A} \omega_A\ d\omega_A &= \int_0^{\theta} \left(2 + 0.006\theta^2\right)\ d\theta \\[1pt] \left[\frac{\left(\omega_A\right)^2}{2}\right]_{15}^{\omega_A} &= \Big[2\theta + 0.002\theta^3\Big]_0^{\theta} \\[1pt] \frac{\left(\omega_A\right)^2}{2} - 112.5 &= 2\theta + 0.002\theta^3 \\[1pt] \frac{\left(\omega_A\right)^2}{2} &= 0.002\theta^3 + 2\theta + 112.5 \\[1pt] \omega_A &= \sqrt{0.004\theta^3 + 4\theta + 225} \\[1pt] \end{aligned}

Angular velocity of gear `A` when `\theta = 10 \times 2\pi = 62.83\text{ rad}`.

\begin{aligned} \omega_A &= \sqrt{0.004\theta^3 + 4\theta + 225} \\[1pt] \omega_A &= \sqrt{0.004 \left(62.83\right)^3 + \left(4 \times 62.83\right) + 225} \\[1pt] \omega_A &= 38.32\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `A` at `\theta = 62.83\text{ rad}`.

\begin{aligned} v_A &= r_A \omega_A \\[1pt] v_A &= 100 \times 38.32 \\[1pt] v_A &= 3832.14\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of gear `B` at `\theta = 62.83\text{ rad}`.

\begin{aligned} \omega_A &= \frac{v_A}{r_B} \\[1pt] \omega_A &= \frac{3832.14}{175} \\[1pt] \omega_A &= 21.9\text{ rad/s} \\[1pt] \end{aligned}

The disk starts at `\omega_0 = 1\text{ rad/s}` when `\theta = 0`, and is given an angular acceleration `\alpha = \left(0.3\theta\right)\text{ rad/s}`, where `\theta` is in radians.

Determine the magnitudes of the normal and tangential components of acceleration of a point `P` on the rim of the disk when `\theta = 1\text{ rev}`.

Angular displacement when `\theta = 1\text{ rev}`.

At `1` revolution, `\theta = 2\pi = 6.28\text{ rad}`.

Angular velocity function.

\begin{aligned} \alpha &= 0.3\theta \\[1pt] \omega\frac{d\omega}{d\theta} &= 0.3\theta \\[1pt] \omega\ d\omega &= 0.3\theta \ d\theta \\[1pt] \int_{1}^{\omega} \omega\ d\omega &= \int_0^{\theta} 0.3\theta \ d\theta \\[1pt] \left[\frac{\omega^2}{2}\right]_{1}^{\omega} &= 0.3\left[\frac{\theta^2}{2}\right]_0^{\theta} \\[1pt] \omega &= \sqrt{0.3\theta^2 + 1} \\[1pt] \end{aligned}

Angular velocity at `\theta = 2\pi = 6.28\text{ rad}`.

\begin{aligned} \omega &= \sqrt{0.3\theta^2 + 1} \\[1pt] \omega &= \sqrt{\left(0.3 \times 6.28^2\right) + 1} \\[1pt] \omega &= 3.58\text{ rad/s} \\[1pt] \end{aligned}

Tangential acceleration.

\begin{aligned} a_t &= r\alpha \\[1pt] a_t &= r \left(0.3\theta\right) \\[1pt] a_t &= 0.4 \left(0.3 \times 6.28\right) \\[1pt] a_t &= 0.75\text{ m/s}^2 \\[1pt] \end{aligned}

Normal acceleration.

\begin{aligned} a_n &= r\omega^2 \\[1pt] a_n &= 0.4 \times 3.58^2 \\[1pt] a_n &= 5.14\text{ rad/s}^2 \\[1pt] \end{aligned}

The disk starts from rest and is given an angular acceleration `\alpha = \left(5t^{1/2}\right)\text{ rad/s}^2`, where `t` is in seconds.

Determine the magnitudes of the normal and tangential components of acceleration of a point `P` on the rim of the disk when `t = 2\text{ s}`.

Angular acceleration.

\begin{aligned} \alpha &= 5t^{1/2} \\[1pt] \alpha &= 5 \times 2^{1/2} \\[1pt] \alpha &= 7.07\text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration.

\begin{aligned} a_t &= r\alpha \\[1pt] a_t &= 0.4 \times 7.07 \\[1pt] a_t &= 2.83\text{ m/s}^2 \\[1pt] \end{aligned}

Angular velocity.

\begin{aligned} \alpha &= 5t^{1/2} \\[1pt] \frac{d\omega}{dt} &= 5t^{1/2} \\[1pt] \int_0^\omega d\omega &= \int_0^2 5t^{1/2} \ dt \\[1pt] \omega &= 5\left[\frac{2}{3}t^{3/2}\right]_0^2 \\[1pt] \omega &= 9.43\text{ rad/s} \\[1pt] \end{aligned}

Normal acceleration.

\begin{aligned} a_n &= r \omega^2\\[1pt] a_n &= 0.4 \times 9.43^2\\[1pt] a_n &= 35.56 \text{ rad/s}^2\\[1pt] \end{aligned}

The disk starts from rest and is given an angular acceleration `\alpha = \left(2t^2\right)\text{ rad/s}^2`, where `t` is in seconds.

Determine the angular velocity of the disk and its angular displacement when `t = 4\text{ s}`.

Angular velocity function.

\begin{aligned} \alpha &= 2t^2 \\[1pt] \frac{d\omega}{dt} &= 2t^2 \\[1pt] \int_0^\omega d\omega &= \int_0^t 2t^2 \ dt \\[1pt] \omega &= 0.67 t^3 \\[1pt] \end{aligned}

Angular velocity at `t = 4\text{ seconds}`.

\begin{aligned} \omega &= 0.67 t^3 \\[1pt] \omega &= 0.67 \times 4^3 \\[1pt] \omega &= 42.67\text{ rad/s} \\[1pt] \end{aligned}

Angular displacement function.

\begin{aligned} \omega &= 0.67 t^3 \\[1pt] \frac{d\theta}{dt} &= 0.67 t^3 \\[1pt] \int_0^\theta d\theta &= \int_0^t 0.67 t^3 \ dt \\[1pt] \theta &= 0.17 t^4\text{ rad} \\[1pt] \end{aligned}

Angular displacement at `t = 4\text{ seconds}`.

\begin{aligned} \theta &= 0.17 t^4\text{ rad} \\[1pt] \theta &= 0.17 \times 4^4\text{ rad} \\[1pt] \theta &= 42.67\text{ rad} \\[1pt] \end{aligned}

The power of a bus engine is transmitted using the belt-and-pulley arrangement shown.

If the engine turns pulley `A` at `\omega_A = 60\text{ rad/s}`, determine the angular velocities of the generator pulley `B` and the air-conditioning pulley `C`.

The hub at `D` is rigidly connected to `B` and turns with it.

Linear velocity of pulley `A`.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 75 \times 60 \\[1pt] v_A &= 4500\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of pulley `B`.

\begin{aligned} \omega_B &= \frac{v_A}{\left(r_B\right)_{\text{inner}}} \\[1pt] \omega_B &= \frac{4500}{25} \\[1pt] \omega_B &= 180\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of pulley `B`.

\begin{aligned} v_B &= \left(r_B\right)_{\text{outer}} \omega_B\\[1pt] v_B &= 100 \times 180 \\[1pt] v_B &= 18000\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of pulley `C`.

\begin{aligned} \omega_C &= \frac{v_B}{r_C} \\[1pt] \omega_C &= \frac{18000}{50} \\[1pt] \omega_C &= 360\text{ rad/s} \\[1pt] \end{aligned}

The power of a bus engine is transmitted using the belt-and-pulley arrangement shown.

If the engine turns pulley `A` at `\omega_A = \left(20t + 40\right)\text{ rad/s}`, where `t` is in seconds, determine the angular velocities of the generator pulley `B` and the air-conditioning pulley `C` when `t = 3\text{ s}`.

Angular velocity of pulley `A` at `t = 3\text{ seconds}`.

\begin{aligned} \omega_A &= 20t + 40 \\[1pt] \omega_A &= \left(20 \times 3\right) + 40 \\[1pt] \omega_A &= 100\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of pulley `A` at `t = 3\text{ seconds}`.

\begin{aligned} v_A &= r_A\omega_A \\[1pt] v_A &= 75 \times 100 \\[1pt] v_A &= 7500\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of pulley `B` at `t = 3\text{ seconds}`.

\begin{aligned} \omega_B &= \frac{v_A}{\left(r_B\right)_{\text{inner}}} \\[1pt] \omega_B &= \frac{7500}{25} \\[1pt] \omega_B &= 300\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of pulley `B` at `t = 3\text{ seconds}`.

\begin{aligned} v_B &= \left(r_B\right)_{\text{outer}} \omega_B\\[1pt] v_B &= 100 \times 300 \\[1pt] v_B &= 30000\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of pulley `C` at `t = 3\text{ seconds}`.

\begin{aligned} \omega_C &= \frac{v_B}{r_C} \\[1pt] \omega_C &= \frac{30000}{50} \\[1pt] \omega_C &= 600\text{ rad/s} \\[1pt] \end{aligned}

The cord, which is wrapped around the disk, is given an acceleration of `a = \left(10t\right)\text{ m/s}^2`, where `t` is in seconds.

Starting from rest, determine the angular displacement, angular velocity, and angular acceleration of the disk when `t = 3\text{ s}`.

Linear acceleration at `t = 3\text{ seconds}`.

\begin{aligned} a &= 10t \\[1pt] a &= 10 \times 3 \\[1pt] a &= 30\text{ m/s}^2 \\[1pt] \end{aligned}

Angular acceleration at `t = 3\text{ seconds}`.

\begin{aligned} \alpha &= \frac{a}{r} \\[1pt] \alpha &= \frac{30}{0.5} \\[1pt] \alpha &= 60\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear velocity function.

\begin{aligned} a &= 10t \\[1pt] \frac{dv}{dt} &= 10t \\[1pt] \int_0^v dv &= \int_0^t 10t \ dt \\[1pt] v &= 5t^2 \\[1pt] \end{aligned}

Linear velocity at `t = 3\text{ seconds}`.

\begin{aligned} v &= 5t^2 \\[1pt] v &= 5 \times 3^2 \\[1pt] v &= 45\text{ m/s} \\[1pt] \end{aligned}

Angular velocity at `t = 3\text{ seconds}`.

\begin{aligned} \omega &= \frac{v}{r} \\[1pt] \omega &= \frac{45}{0.5} \\[1pt] \omega &= 90\text{ rad/s} \\[1pt] \end{aligned}

Linear displacement function.

\begin{aligned} v &= 5t^2 \\[1pt] \frac{dS}{dt} &= 5t^2 \\[1pt] \int_0^S dS &= \int_0^t 5t^2 \ dt \\[1pt] S &= 1.6667t^3 \\[1pt] \end{aligned}

Linear displacement at `t = 3\text{ seconds}`.

\begin{aligned} S &= 1.6667t^3 \\[1pt] S &= 1.6667 \times 3^3 \\[1pt] S &= 45\text{ m} \\[1pt] \end{aligned}

Angular displacement at `t = 3\text{ seconds}`.

\begin{aligned} \theta &= \frac{S}{r} \\[1pt] \theta &= \frac{45}{0.5} \\[1pt] \theta &= 90\text{ rad} \\[1pt] \end{aligned}

At the instant `\omega_A = 5\text{ rad/s}`, pulley `A` is given a constant angular acceleration `\alpha_A = 6\text{ rad/s}^2`.

Determine the magnitude of acceleration of point `B` on pulley `C` when `A` rotates `2` revolutions.

Pulley `C` has an inner hub which is fixed to its outer one and turns with it.

Angular velocity of pulley `A`.

\begin{aligned} \left(\omega_A\right)^2 &= \left(\omega_0\right)_A^2 + 2\alpha_A\theta_A \\[1pt] \left(\omega_A\right)^2 &= \left(5\right)_A^2 + 2 \left(6\right)\left(12.57\right) \\[1pt] \omega_A &= 13.26\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of pulley `A`.

\begin{aligned} v_A &= r_A \omega_A \\[1pt] v_A &= \left(0.05 \text{ meters}\right) \times 13.26 \\[1pt] v_A &= 0.66\text{ m/s} \\[1pt] \end{aligned}

Linear acceleration of pulley `A`.

\begin{aligned} a_A &= r_A \alpha_A \\[1pt] a_A &= \left(0.05 \text{ meters}\right) \times 6 \\[1pt] a_A &= 0.3\text{ m/s} \\[1pt] \end{aligned}

Pulley `C` is attached pulley `A` using the same belt, therefore, the linear velocity and acceleration of both pulleys is the same.

\begin{aligned} \omega_C &= \frac{v_A}{r_C} \\[1pt] \omega_C &= \frac{0.66}{0.04} \\[1pt] \omega_C &= 16.57\text{ rad/s} \\[1pt] \end{aligned}

Tangential acceleration at point `B`.

\begin{aligned} \left(a_t\right)_B &= r_B \alpha_C \\[1pt] \left(a_t\right)_B &= 0.06 \times 7.5 \\[1pt] \left(a_t\right)_B &= 0.45\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal acceleration at point `B`.

\begin{aligned} \left(a_c\right)_B &= \frac{\left(v_B\right)^2}{r_B} \\[1pt] \left(a_c\right)_B &= \frac{\left(r_B\omega_C\right)^2}{r_B} \\[1pt] \left(a_c\right)_B &= r_B\left(\omega_C\right)^2 \\[1pt] \left(a_c\right)_B &= 0.06\left(16.57\right)^2 \\[1pt] \left(a_c\right)_B &= 16.48\text{ m/s}^2 \\[1pt] \end{aligned}

Net acceleration at point `B`.

\begin{aligned} \left(a_\text{net}\right)_B &= \sqrt{\left(a_t\right)_B^2 + \left(a_c\right)_B^2} \\[1pt] \left(a_\text{net}\right)_B &= \sqrt{0.45^2 + 16.48^2} \\[1pt] \left(a_\text{net}\right)_B &= 16.49\text{ m/s}^2 \\[1pt] \end{aligned}

At the instant `\omega_A = 5\text{ rad/s}`, pulley `A` is given an angular acceleration `\alpha = \left(0.8\theta\right)\text{ rad/s}^2`, where `\theta` is in radians.

Determine the magnitude of acceleration of point `B` on pulley `C` when `A` rotates `3` revolutions.

Pulley `C` has an inner hub which is fixed to its outer one and turns with it.

Angular velocity function of pulley `A`.

\begin{aligned} \alpha &= 0.8\theta \\[1pt] \omega\frac{d\omega}{d\theta} &= 0.8\theta \\[1pt] \int_5^{\omega_A} \omega\ d\omega &= \int_0^\theta 0.8\theta\ d\theta \\[1pt] \left[\frac{\omega^2}{2}\right]_5^{\omega_A} &= 0.8\left[\frac{\theta^2}{2}\right]_0^\theta \\[1pt] \frac{\left(\omega_A\right)^2}{2} - 12.5 &= 0.4\theta^2 \\[1pt] \frac{\left(\omega_A\right)^2}{2} &= 0.4\theta^2 + 12.5 \\[1pt] \omega_A &= \sqrt{0.8\theta^2 + 25} \\[1pt] \end{aligned}

`\omega_A` is the angular velocity of pulley `A` after it rotates `3` revolutions.

\begin{aligned} \omega_A &= \sqrt{0.8\theta^2 + 25} \\[1pt] \omega_A &= \sqrt{0.8 \left(6\pi\right)^2 + 25} \\[1pt] \omega_A &= 17.59\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of pulley `A`.

\begin{aligned} v_A &= r_A \omega_A \\[1pt] v_A &= \left(0.05 \text{ meters}\right) \times 17.59 \\[1pt] v_A &= 0.88\text{ m/s} \\[1pt] \end{aligned}

Angular acceleration of pulley `A`.

\begin{aligned} \alpha &= 0.8\theta \\[1pt] \alpha_A &= 0.8 \times 18.85 \\[1pt] \alpha_A &= 15.08\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear acceleration of pulley `A`.

\begin{aligned} a_A &= r_A \alpha_A \\[1pt] a_A &= \left(0.05 \text{ meters}\right) \times 15.08 \\[1pt] a_A &= 0.75\text{ m/s} \\[1pt] \end{aligned}

Pulley `C` is attached pulley `A` using the same belt, therefore, the linear velocity and acceleration of both pulleys is the same.

\begin{aligned} \omega_C &= \frac{v_A}{r_C} \\[1pt] \omega_C &= \frac{0.88}{0.04} \\[1pt] \omega_C &= 21.98\text{ rad/s} \\[1pt] \end{aligned}

Tangential acceleration at point `B`.

\begin{aligned} \left(a_t\right)_B &= r_B \alpha_C \\[1pt] \left(a_t\right)_B &= 0.06 \times 18.85 \\[1pt] \left(a_t\right)_B &= 1.13\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal acceleration at point `B`.

\begin{aligned} \left(a_c\right)_B &= \frac{\left(v_B\right)^2}{r_B} \\[1pt] \left(a_c\right)_B &= \frac{\left(r_B\omega_C\right)^2}{r_B} \\[1pt] \left(a_c\right)_B &= r_B\left(\omega_C\right)^2 \\[1pt] \left(a_c\right)_B &= 0.06\left(21.98\right)^2 \\[1pt] \left(a_c\right)_B &= 28.99\text{ m/s}^2 \\[1pt] \end{aligned}

Net acceleration at point `B`.

\begin{aligned} \left(a_\text{net}\right)_B &= \sqrt{\left(a_t\right)_B^2 + \left(a_c\right)_B^2} \\[1pt] \left(a_\text{net}\right)_B &= \sqrt{1.13^2 + 28.99^2} \\[1pt] \left(a_\text{net}\right)_B &= 29.01\text{ m/s}^2 \\[1pt] \end{aligned}

If gear `A` rotates with an angular velocity of `\omega_A = \left(\theta_A + 1\right)\text{ rad/s}`, where `\theta_A` is the angular displacement of gear `A`, measured in radians, determine the angular acceleration of gear `D` when `\theta_A = 3\text{ rad}`, starting from rest.

Gears `A, B, C,` and `D` have radii of `15\text{ mm}, 50\text{ mm}, 25\text{ mm},` and `75\text{ mm}`, respectively.

Angular acceleration of gear `A` when `\theta_A = 3\text{ rad}`.

\begin{aligned} \omega_A &= \theta_A + 1 \\[1pt] \frac{d\omega_A}{dt} &= \frac{d}{dt}\Big[\theta_A + 1\Big] \\[1pt] \alpha_A &= \omega_A \\[1pt] \alpha_A &= \theta_A + 1 \color{green}\text{ //substituted } \omega_A = \theta_A + 1 \\[1pt] \alpha_A &= 3 + 1 \\[1pt] \alpha_A &= 4\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear acceleration of gear `A`.

\begin{aligned} a_A &= r_A\alpha_A \\[1pt] a_A &= 15 \times 4 \\[1pt] a_A &= 60\text{ mm/s}^2 \\[1pt] \end{aligned}

Angular accelerations of gears `B` and `C` are equal because they belong to the same longitudinal axis.

\begin{aligned} \alpha_B = \alpha_C &= \frac{a_A}{r_B} \\[1pt] \alpha_B = \alpha_C &= \frac{60}{50} \\[1pt] \alpha_B = \alpha_C &= 1.2\text{ rad/s}^2 \\[1pt] \end{aligned}

Linear acceleration of gear `C`.

\begin{aligned} a_C &= r_C\alpha_C \\[1pt] a_C &= 25 \times 1.2 \\[1pt] a_C &= 30\text{ mm/s}^2 \\[1pt] \end{aligned}

Angular acceleration of gear `D`.

\begin{aligned} \alpha_D &= \frac{a_C}{r_D} \\[1pt] \alpha_D &= \frac{30}{75} \\[1pt] \alpha_D &= 0.4\text{ rad/s}^2 \\[1pt] \end{aligned}

If gear `A` rotates with a constant angular acceleration of `\alpha_A = 90\text{ rad/s}^2`, starting from rest, determine the time required for gear `D` to attain an angular velocity of `600\text{ rpm}`.

Also, find the number of revolutions of gear `D` to attain this angular velocity.

Gears `A, B, C,` and `D` have radii of `15\text{ mm}, 50\text{ mm}, 25\text{ mm},` and `75\text{ mm}`, respectively.

Convert `\omega_D` from `\text{rpm}` to `\text{rad/s}`.

\begin{aligned} \omega_D &= \frac{600 \text{ rpm}}{60\text{ seconds}} \times 2\pi \\[1pt] \omega_D &= 62.83\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `D`.

\begin{aligned} v_D &= r_D\omega_D \\[1pt] v_D &= 75 \times 62.83 \\[1pt] v_D &= 4712.39\text{ mm/s} \\[1pt] \end{aligned}

Angular velocities of gears `C` and `B` are equal because they belong to the same longitudinal axis.

\begin{aligned} \omega_C = \omega_B &= \frac{v_D}{r_C} \\[1pt] \omega_C = \omega_B &= \frac{4712.39}{25} \\[1pt] \omega_C = \omega_B &= 188.5\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of gear `B`.

\begin{aligned} v_B &= r_B\omega_B \\[1pt] v_B &= 50 \times 188.5 \\[1pt] v_B &= 9424.78\text{ mm/s} \\[1pt] \end{aligned}

Angular velocity of gear `A`.

\begin{aligned} \omega_A &= \frac{v_B}{r_A} \\[1pt] \omega_A &= \frac{9424.78}{15} \\[1pt] \omega_A &= 628.32\text{ rad/s} \\[1pt] \end{aligned}

Now that the angular velocity of gear `A` is known, find the time taken for it to reach that velocity.

\begin{aligned} \omega_A &= \left(\omega_0\right)_A + \alpha_A t \\[1pt] 628.32 &= 0 + 90 t \\[1pt] t &= \frac{628.32}{90} \\[1pt] t &= 6.98\text{ seconds} \\[1pt] \end{aligned}

Find the angular acceleration of gear `D` in `6.98\text{ seconds}`.

\begin{aligned} \omega_D &= \left(\omega_0\right)_D + \alpha_D t \\[1pt] 62.83 &= 0 + \alpha_D 6.98 \\[1pt] \alpha_D &= \frac{62.83}{6.98} \\[1pt] \alpha_D &= 9\text{ rad/s}^2 \\[1pt] \end{aligned}

Angular displacement of gear `D`.

\begin{aligned} \theta_D &= \left(\omega_0\right)_D t + \frac{1}{2}\alpha_D t^2 \\[1pt] \theta_D &= \left(0\right)t + \frac{1}{2} \left(9\right) 6.98^2 \\[1pt] \theta_D &= 219.32\text{ rad} \\[1pt] \end{aligned}

A wheel has an initial clockwise angular velocity of `10\text{ rad/s}` and a constant angular acceleration of `3\text{ rad/s}^2`.

Determine the number of revolutions it must undergo to acquire a clockwise angular velocity of `15\text{ rad/s}`.

What time is required?

`\theta` is the angular displacement of the wheel when it acquires an angular velocity of `15\text{ rad/s}`.

\begin{aligned} \omega^2 &= \omega_0^2 + 2\alpha\theta \\[1pt] 15^2 &= 10^2 + 2 \left(3\right)\theta \\[1pt] \theta &= 20.83\text{ rad} \\[1pt] \end{aligned}

The disk is driven by a motor such that the angular position of the disk is defined by `\theta = \left(20t + 4t^2\right)\text{ rad}`, where `t` is in seconds.

Determine the number of revolutions, the angular velocity, and angular acceleration of the disk when `t = 90\text{ s}`.

Angular displacement.

\begin{aligned} \theta &= 20t + 4t^2 \\[1pt] \theta &= \left(20 \times 90\right) + \left(4 \times 90^2\right) \\[1pt] \theta &= 34200\text{ rad} \\[1pt] \end{aligned}

Angular velocity.

\begin{aligned} \theta &= 20t + 4t^2 \\[1pt] \frac{d\theta}{dt} &= \frac{d}{dt}\Big[20t + 4t^2\Big] \\[1pt] \omega &= 20 + 8t \\[1pt] \omega &= 740\text{ rad/s} \\[1pt] \end{aligned}

Angular acceleration.

\begin{aligned} \omega &= 20 + 8t \\[1pt] \frac{d\omega}{dt} &= \frac{d}{dt}\Big[20 + 8t\Big] \\[1pt] \alpha &= 8\text{ rad/s}^2 \\[1pt] \end{aligned}

The disk is originally rotating at `\omega_0 = 12\text{ rad/s}`.

If it is subjected to a constant angular acceleration of `\alpha = 20\text{ rad/s}^2`, determine the magnitudes of the velocity and the `n` and `t` components of acceleration of point `B` when the disk undergoes `2` revolutions.

Angular velocity of the disk after `2` revolutions.

\begin{aligned} \omega^2 &= \left(\omega_0\right)^2 + 2\alpha\theta \\[1pt] \omega &= \left(12\right)^2 + \left(2 \times 20 \times 4\pi\right) \\[1pt] \omega &= 25.43\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of point `B` after `2` revolutions.

\begin{aligned} v_B &= r_B\omega \\[1pt] v_B &= 0.4 \times 25.43 \\[1pt] v_B &= 10.17\text{ m/s} \\[1pt] \end{aligned}

Tangential acceleration of point `B` after `2` revolutions.

\begin{aligned} \left(a_t\right)_B &= r_B\alpha \\[1pt] \left(a_t\right)_B &= 0.4 \times 20 \\[1pt] \left(a_t\right)_B &= 8\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal/Normal acceleration of point `B` after `2` revolutions.

\begin{aligned} \left(a_c\right)_B &= \frac{\left(v_B\right)^2}{r_B} \\[1pt] \left(a_c\right)_B &= \frac{\left(10.17\right)^2}{0.4} \\[1pt] \left(a_c\right)_B &= 258.66\text{ m/s}^2 \\[1pt] \end{aligned}

The disk is originally rotating at `\omega_0 = 12\text{ rad/s}`.

If it is subjected to a constant angular acceleration of `\alpha = 20\text{ rad/s}^2`, determine the magnitudes of the velocity and the `n` and `t` components of acceleration of point `A` at the instant `t = 2\text{ s}`.

Angular velocity of the disk.

\begin{aligned} \omega &= \omega_0 + \alpha t \\[1pt] \omega &= 12 + \left(20 \times 2\right) \\[1pt] \omega &= 52\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of point `A`.

\begin{aligned} v_A &= r_A\omega \\[1pt] v_A &= 0.5 \times 52 \\[1pt] v_A &= 26\text{ m/s} \\[1pt] \end{aligned}

Tangential acceleration of point `A`.

\begin{aligned} \left(a_t\right)_A &= r_A\alpha \\[1pt] \left(a_t\right)_A &= 0.5 \times 20 \\[1pt] \left(a_t\right)_A &= 10\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal/Normal acceleration of point `A`.

\begin{aligned} \left(a_c\right)_A &= \frac{\left(v_A\right)^2}{r_A} \\[1pt] \left(a_c\right)_A &= \frac{\left(26\right)^2}{0.5} \\[1pt] \left(a_c\right)_A &= 1352\text{ m/s}^2 \\[1pt] \end{aligned}

The angular acceleration of the disk is defined by `\alpha = \left(3t^2 + 12\right)\text{ rad/s}`, where `t` is in seconds.

If the disk is originally rotating at `\omega_0 = 12\text{ rad/s}`, determine the magnitude of the velocity and the `n` and `t` components of acceleration of point `A` on the disk when `t = 2\text{ s}`.

Angular velocity function.

\begin{aligned} \alpha &= 3t^2 + 12 \\[1pt] \frac{d\omega}{dt} &= 3t^2 + 12 \\[1pt] d\omega &= \left(3t^2 + 12\right)\ dt \\[1pt] \int_{12}^{\omega} d\omega &= \int_0^t \left(3t^2 + 12\right)\ dt \\[1pt] \omega - 12 &= t^3 + 12t \\[1pt] \omega &= t^3 + 12t + 12 \\[1pt] \end{aligned}

Angular velocity at `t = 2\text{ seconds}`.

\begin{aligned} \omega &= t^3 + 12t + 12 \\[1pt] \omega &= 2^3 + \left(12 \times 2\right) + 12 \\[1pt] \omega &= 44\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of point `A` at `t = 2\text{ seconds}`.

\begin{aligned} v_A &= r_A\omega \\[1pt] v_A &= 0.5 \times 44 \\[1pt] v_A &= 22\text{ m/s} \\[1pt] \end{aligned}

Angular acceleration at `t = 2\text{ seconds}`.

\begin{aligned} \alpha &= 3t^2 + 12 \\[1pt] \alpha &= \left(3 \times 2^2\right) + 12 \\[1pt] \alpha &= 24\text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration of point `A` at `t = 2\text{ seconds}`.

\begin{aligned} \left(a_t\right)_A &= r_A\alpha \\[1pt] \left(a_t\right)_A &= 0.5 \times 24 \\[1pt] \left(a_t\right)_A &= 12\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal/Normal acceleration of point `A` at `t = 2\text{ seconds}`.

\begin{aligned} \left(a_c\right)_A &= \frac{\left(v_A\right)^2}{r_A} \\[1pt] \left(a_c\right)_A &= \frac{\left(22\right)^2}{0.5} \\[1pt] \left(a_c\right)_A &= 968\text{ m/s}^2 \\[1pt] \end{aligned}

The angular velocity of the disk is defined by `\omega = \left(5t^2 + 2\right)\text{ rad/s}`, where `t` is in seconds.

Determine the magnitudes of the velocity and acceleration of point `A` on the disk when `t = 0.5\text{ s}`.

Angular velocity at `t = 0.5\text{ seconds}`.

\begin{aligned} \omega &= 5t^2 + 2 \\[1pt] \omega &= 5 \left(0.5^2\right) + 2 \\[1pt] \omega &= 3.25\text{ rad/s} \\[1pt] \end{aligned}

Linear velocity of point `A` at `t = 0.5\text{ seconds}`.

\begin{aligned} v &= r\omega \\[1pt] v &= 0.8 \times 3.25 \\[1pt] v &= 2.6 \text{ m/s} \\[1pt] \end{aligned}

Angular acceleration function.

\begin{aligned} \omega &= 5t^2 + 2 \\[1pt] \frac{d\omega}{dt} &= \frac{d}{dt}\Big[5t^2 + 2\Big] \\[1pt] \alpha &= 10t \\[1pt] \end{aligned}

Angular acceleration at `t = 0.5\text{ seconds}`.

\begin{aligned} \alpha &= 10t \\[1pt] \alpha &= 10 \times 0.5 \\[1pt] \alpha &= 5\text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration of point `A` at `t = 0.5\text{ seconds}`.

\begin{aligned} a_t &= r\alpha \\[1pt] a_t &= 0.8 \times 5 \\[1pt] a_t &= 4\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal acceleration of point `A` at `t = 0.5\text{ seconds}`.

\begin{aligned} a_c &= \frac{v^2}{r} \\[1pt] a_c &= \frac{2.6^2}{0.8} \\[1pt] a_c &= 8.45\text{ m/s}^2 \\[1pt] \end{aligned}

Net acceleration of point `A` at `t = 0.5\text{ seconds}`.

\begin{aligned} a_{\text{net}} &= \sqrt{\left(a_t\right)^2 + \left(a_c\right)^2} \\[1pt] a_{\text{net}} &= \sqrt{\left(4\right)^2 + \left(8.45\right)^2} \\[1pt] a_{\text{net}} &= 9.35\text{ m/s}^2 \\[1pt] \end{aligned}

For a short period of time, the motor turns gear `A` with a constant angular acceleration of `\alpha_A = 4.5\text{ rad/s}^2`, starting from rest.

Determine the velocity of the cylinder and the distance it travels in three seconds.

The cord is wrapped around pulley `D` which is rigidly attached to gear `B`.

Find the tangential acceleration at point `P`.

\begin{aligned} a &= r \alpha \\[1pt] a_P &= r_A\alpha_A \\[1pt] a_P &= 75 \times 4.5 \\[1pt] a_P &= 337.5 \text{ mm/s}^2 \\[1pt] \end{aligned}

Note, point `P` on Gear `B` has the same velocity at point `P` on Gear `A`. Using that, find the angular velocity of Gear `B`.

\begin{aligned} a &= r \alpha \\[1pt] a_P &= r_B \alpha_B \\[1pt] 337.5 &= 225 \alpha_B \\[1pt] \alpha_B &= 1.5\text{ rad/s}^2 \\[1pt] \end{aligned}

`\left(\omega_B\right)_{\text{final}}` is the velocity of gear `B` at `t = 3\text{ seconds}`.

\begin{aligned} \left(\omega_B\right)_{\text{final}} &= \left(\omega_B\right)_{\text{initial}} + \alpha_B t \\[1pt] \left(\omega_B\right)_{\text{final}} &= 0 + \left(1.5 \times 3\right) \\[1pt] \left(\omega_B\right)_{\text{final}} &= 4.5\text{ rad/s} \\[1pt] \end{aligned}

The velocity of the cylinder is equal to the velocity of the point `P'`.

\begin{aligned} v &= r \omega \\[1pt] v_P' &= r_D \omega_B \\[1pt] v_P' &= 125 \times 4.5 \\[1pt] v_P' &= 562.5\text{ mm/s} \\[1pt] \end{aligned}

`\theta_B` is the angular displacement of gear `B` at `t = 3\text{ seconds}`.

\begin{aligned} \theta_B &= \left(\omega_B\right)_{\text{initial}}t + \frac{1}{2}\alpha_B t^2 \\[1pt] \theta_B &= 0t + \frac{1}{2}\left(1.5\right) 3^2 \\[1pt] \theta_B &= 6.75\text{ rad/s} \\[1pt] \end{aligned}

`S_P'` is the distance travelled by the cylinder at `t = 3\text{ seconds}`.

\begin{aligned} S &= r\theta \\[1pt] S_P' &= r_D\theta_B \\[1pt] S_P' &= 125 \times 6.75 \\[1pt] S_P' &= 843.75\text{ mm} \\[1pt] \end{aligned}

A wheel has an angular acceleration of `\alpha = \left(0.5 \theta\right)\text{ rad/s}^2`, where `\theta` is in radians.

Determine the magnitude of the velocity and acceleration of a point `P` located on its rim after the wheel has rotated `2` revolutions.

The wheel has a radius of `0.2\text{ m}` and starts at `\omega_0 = 2\text{ rad/s}`.

Angular velocity function of the wheel.

\begin{aligned} \alpha &= 0.5 \theta \\[1pt] \omega \frac{d\omega}{d\theta} &= 0.5 \theta \\[1pt] \omega\ d\omega &= 0.5 \theta \ d\theta \\[1pt] \int_2^\omega \omega\ d\omega &= \int_0^\theta 0.5 \theta \ d\theta \\[1pt] \left[\frac{\omega^2}{2}\right]_2^\omega &= 0.5 \left[\frac{\theta^2}{2}\right]_0^\theta \\[1pt] \frac{\omega^2}{2} - 2 &= 0.25 \theta^2 \\[1pt] \omega &= \sqrt{0.5 \theta^2 + 4} \\[1pt] \end{aligned}

Angular velocity of the wheel after it has rotated `2` revolutions.

\begin{aligned} \omega &= \sqrt{0.5 \theta^2 + 4} \\[1pt] \omega &= \sqrt{0.5 \left(4\pi\right)^2 + 4} \\[1pt] \omega &= 9.11\text{ rad/s} \\[1pt] \end{aligned}

Velocity of the point `P` after the wheel has rotated `2` revolutions.

\begin{aligned} v &= r\omega \\[1pt] v &= 0.2 \times 9.11 \\[1pt] v &= 1.82\text{ m/s} \\[1pt] \end{aligned}

Angular acceleration of the wheel after it has rotated `2` revolutions.

\begin{aligned} \alpha &= 0.5 \theta \\[1pt] \alpha &= 0.5 \left(4\pi\right) \\[1pt] \alpha &= 6.28\text{ rad/s}^2 \\[1pt] \end{aligned}

Tangential acceleration of the point `P` after the wheel has rotated `2` revolutions.

\begin{aligned} \left(a_t\right)_P &= r\alpha \\[1pt] \left(a_t\right)_P &= 0.2 \times 6.28 \\[1pt] \left(a_t\right)_P &= 1.26\text{ m/s}^2 \\[1pt] \end{aligned}

Centripetal acceleration of the point `P` after the wheel has rotated `2` revolutions.

\begin{aligned} \left(a_c\right)_P &= \frac{v^2}{r} \\[1pt] \left(a_c\right)_P &= \frac{\left(r\omega\right)^2}{r} \\[1pt] \left(a_c\right)_P &= \frac{r^2\omega^2}{r} \\[1pt] \left(a_c\right)_P &= r\omega^2 \\[1pt] \left(a_c\right)_P &= 0.2 \times 9.11^2 \\[1pt] \left(a_c\right)_P &= 16.59\text{ m/s}^2 \\[1pt] \end{aligned}

Net acceleration of the point `P` after the wheel has rotated `2` revolutions.

\begin{aligned} \left(a_{\text{net}}\right)_P &= \sqrt{\left(a_t\right)_P^2 + \left(a_c\right)_P^2} \\[1pt] \left(a_{\text{net}}\right)_P &= \sqrt{\left(1.26\right)^2 + \left(16.59\right)^2} \\[1pt] \left(a_{\text{net}}\right)_P &= 16.64\text{ m/s}^2 \\[1pt] \end{aligned}

The bucket is hoisted by the rope that wraps around a drum wheel.

If the angular displacement of the wheel is `\theta = \left(0.5t^3 + 15t\right)\text{ rad}`, where `t` is in seconds, determine the velocity and acceleration of the bucket when `t = 3\text{ s}`.

Angular velocity function of the wheel.

\begin{aligned} \theta &= 0.5t^3 + 15t \\[1pt] \frac{d\theta}{dt} &= \frac{d}{dt}\Big[0.5t^3 + 15t\Big] \\[1pt] \omega &= 1.5t^2 + 15 \\[1pt] \end{aligned}

Angular velocity of the wheel when `t = 3\text{ seconds}`.

\begin{aligned} \omega &= 1.5t^2 + 15 \\[1pt] \omega &= 1.5 \left(3\right)^2 + 15 \\[1pt] \omega &= 28.5\text{ rad/s} \\[1pt] \end{aligned}

Velocity of the bucket when `t = 3\text{ seconds}`.

\begin{aligned} v &= r\omega \\[1pt] v &= 0.75 \times 28.5 \\[1pt] v &= 21.375\text{ ft/s} \\[1pt] \end{aligned}

Angular acceleration function of the wheel.

\begin{aligned} \omega &= 1.5t^2 + 15 \\[1pt] \frac{d\omega}{dt} &= \frac{d}{dt}\Big[1.5t^2 + 15\Big] \\[1pt] \alpha &= 3t \\[1pt] \end{aligned}

Angular acceleration of the wheel when `t = 3\text{ seconds}`.

\begin{aligned} \alpha &= 3t \\[1pt] \alpha &= 3 \times 3 \\[1pt] \alpha &= 9\text{ rad/s}^2 \\[1pt] \end{aligned}

Acceleration of the bucket when `t = 3\text{ seconds}`.

\begin{aligned} a &= r\alpha \\[1pt] a &= 0.75 \times 9 \\[1pt] a &= 6.75\text{ ft/s}^2 \\[1pt] \end{aligned}

The flywheel rotates with an angular velocity of `\omega = \left(4 \theta^{1/2}\right)\text{ rad/s}`, where `\theta` is in radians.

Determine the time it takes to achieve an angular velocity of `\omega = 150\text{ rad/s}`.

When `t = 0`, `\theta = 1\text{ rad}`.

Initial angular velocity.

\begin{aligned} \omega &= 4\theta^{1/2} \\[1pt] \omega_{\text{initial}} &= 4 \left(1\right)^{1/2} \\[1pt] \omega_{\text{initial}} &= 4 \text{ rad/s} \\[1pt] \end{aligned}

`\theta` is the angular displacement when `\omega = 150\text{ rad/s}`.

\begin{aligned} \omega &= 4\theta^{1/2} \\[1pt] 150 &= 4\sqrt\theta \\[1pt] \sqrt\theta &= \frac{150}{4} \\[1pt] \theta &= 1406.25\text{ rad} \\[1pt] \end{aligned}

From the function `\omega = 4\theta^{1/2}` it is known that the angular acceleration is constant.

\begin{aligned} \omega &= 4\theta^{1/2} \\[1pt] \frac{d\omega}{dt} &= \frac{d}{dt}\Big[4\theta^{1/2}\Big] \\[1pt] \alpha &= 4\frac{d}{dt}\Big[\theta^{1/2}\Big] \\[1pt] \alpha &= 2\theta^{-1/2}\omega \\[1pt] \alpha &= 2 \left(1\right)^{-1/2} 4 \\[1pt] \alpha &= 8\text{ rad/s}^2 \\[1pt] \end{aligned}

The flywheel rotates with an angular velocity of `\omega = \left(0.005\theta^2\right)\text{ rad/s}`, where `\theta` is in radians.

Determine the angular acceleration when it has rotated `20\text{ revolutions}`.

When the gear rotates `20\text{ revolutions}`, it achieves an angular velocity of `\omega = 30\text{ rad/s}`, starting from rest.

Determine its constant angular acceleration and the time required.

`\theta` is the angular displacement at the `20^{\text{th}}\text{ revolution}`.

\begin{aligned} \theta &= 20 \times 2\pi \text{ rad} \\[1pt] \hline \theta &= 40\pi \text{ rad} \\[1pt] \end{aligned}

`\alpha` is the angular acceleration at the `20^{\text{th}}\text{ revolution}`.

\begin{aligned} \omega_{\text{final}}^2 &= \omega_{\text{initial}}^2 + 2\alpha\theta \\[1pt] 30^2 &= 0^2 + 2\alpha 40\pi \\[1pt] 30^2 &= 2\alpha 40\pi \\[1pt] \hline \alpha &= 3.58\text{ rad/s}^2 \\[1pt] \end{aligned}

`t` is the time taken to complete `20\text{ revolutions}`.

\begin{aligned} v &= u + at \\[1pt] \omega &= \omega_0 + \alpha t \\[1pt] 30 &= 0 + 3.58 t \\[1pt] \hline t &= 8.38\text{ seconds} \\[1pt] \end{aligned}

Relationship between linear and angular velocity.

\begin{aligned} v &= \frac{dS}{dt} \\[1pt] v &= \frac{d}{dt} \Big[r\theta\Big] \color{green}\text{ //substituted } S = r\theta \\[1pt] v &= r\frac{d\theta}{dt} \\[1pt] v &= r\omega \\[1pt] \end{aligned}

Relationship between linear and angular acceleration.

\begin{aligned} v &= r\omega \\[1pt] \frac{dv}{dt} &= \frac{d}{dt}\Big[r\omega\Big] \\[1pt] a_t &= r\frac{d}{dt}\Big[\omega\Big] \\[1pt] a_t &= r\frac{d\omega}{dt} \\[1pt] a_t &= r\alpha \\[1pt] \end{aligned}

Relationship between centripetal acceleration and angular velocity.

\begin{aligned} a_c &= \frac{v^2}{r} \\[1pt] a_c &= \frac{\left(r\omega\right)^2}{r} \\[1pt] a_c &= \frac{r^\cancel{2}\omega^2}{\cancel{r}} \\[1pt] a_c &= r\omega^2 \\[1pt] \end{aligned}