Question

At time t=0 a grinding wheel has an angular velocity of 24.0 rad/s. It has a constant angular acceleration of 35.0 rad/s^2 until a circuit breaker trips at time t= 2.50 s. From then on, the wheel turns through an angle of 440 rad as it coasts to a stop at constant angular deceleration.

a.) Through what total angle did the wheel turn between t= 0 and the time it stopped? answer in rad

b.) At what time does the wheel stop? (secs)

c.)What was the wheel's angular acceleration as it slowed down?
Express your answer in radians per second per second.

Answer 1

Concepts and reason

The concept required to solve this problem is rotational kinematic equations.

Initially, find the angular displacement of the wheel. Later, find the angular velocity of the wheel.

Finally, find the angular acceleration of the wheel.

Fundamentals

The expression for the angular displacement in rotational kinematics equations is as follows:

${\theta _1} = {\omega _0}t + \frac{1}{2}\alpha {t^2}$

Here, ${\omega _0}$ is the initial angular speed, t is the time taken, and $\alpha$ is the angular acceleration.

The expression for the angular velocity of the wheel is as follows:

$\omega = {\omega _0} + \alpha t$

Here, ${\omega _0}$ is the initial angular speed, t is the time taken, and $\alpha$ is the angular acceleration.

And, the expression for the angular acceleration of the wheel is as follows:

$\alpha ' = \frac{{\omega ' - \omega }}{{\Delta t}}$

Here, $\omega '$ is the angular velocity when the wheel stops and $\Delta t$ is the change in the time.

(a)

The initial angular position is equal to ${\theta _0} = 0^\circ$ .

Now, substitute 24.0 rad/s for ${\omega _0}$ , 2.50 s for t, and $35.0{\rm{ rad/}}{{\rm{s}}^2}$ for $\alpha$ in the equation ${\theta _1} = {\omega _0}t + \frac{1}{2}\alpha {t^2}$ .

$\begin{array}{c}\\{\theta _1} = \left( {24.0{\rm{ rad/s}}} \right)\left( {2.50{\rm{ s}}} \right) + \frac{1}{2}\left( {35.0{\rm{ rad/}}{{\rm{s}}^2}} \right){\left( {2.50{\rm{ s}}} \right)^2}\\\\ = 169.4{\rm{ rad}}\\\end{array}$

Solve for ${\theta _1}$ .

${\theta _1} = 169.4{\rm{ rad}}$

Therefore, the total displacement is as follows:

$\begin{array}{c}\\169.4{\rm{ rad}} + {\rm{440 rad}} = 609.4{\rm{ rad}}\\\\ = {\rm{609 rad}}\\\end{array}$

(b)

Substitute 24.0 rad/s for ${\omega _0}$ , 2.50 s for t, and $35.0{\rm{ rad/}}{{\rm{s}}^2}$ for $\alpha$ in the equation $\omega = {\omega _0} + \alpha t$ .

$\begin{array}{c}\\\omega = \left( {24.0{\rm{ rad/s}}} \right) + \left( {35.0{\rm{ rad/}}{{\rm{s}}^2}} \right)\left( {2.50{\rm{ s}}} \right)\\\\ = 111.5{\rm{ rad/s}}\\\end{array}$

Now, find the time interval by rearranging the equation $\Delta \theta = \frac{1}{2}\left( {\omega + \omega '} \right)\Delta t$ .

$\Delta t = \frac{{2\Delta \theta }}{{\left( {\omega + \omega '} \right)}}$

Substitute 440 rad for $\Delta \theta$ , 111.5 rad/s for $\omega$ , and 0.00 rad/s for $\omega '$ .

$\begin{array}{c}\\\Delta t = \frac{{2\left( {440{\rm{ rad}}} \right)}}{{\left( {111.5{\rm{ rad/s}} + 0.00{\rm{ rad/s}}} \right)}}\\\\ = \frac{{880{\rm{ rad/s}}}}{{111.5{\rm{ rad/s}}}}\\\\ = 7.90{\rm{ s}}\\\end{array}$

Therefore, the total time taken by the wheel to stop is as follows:

$\begin{array}{c}\\t' = 7.90{\rm{ s}} + 2.50{\rm{ s}}\\\\{\rm{ = 10}}{\rm{.4 s}}\\\end{array}$

(c)

Substitute 0.00 rad/s for $\omega '$ , 111.5 rad/s for $\omega$ , and 7.90 s for $\Delta t$ in the equation $\alpha ' = \frac{{\omega ' - \omega }}{{\Delta t}}$ .

$\begin{array}{c}\\\alpha ' = \frac{{0.00{\rm{ rad/s}} - 111.5{\rm{ rad/s}}}}{{7.90{\rm{ s}}}}\\\\ = \frac{{ - 111.5{\rm{ rad/s}}}}{{7.90{\rm{ s}}}}\\\\ = - 14.11{\rm{ rad/}}{{\rm{s}}^2}\\\end{array}$

[Part c]

Ans: Part a

The total angle that the wheel turns between $t = 0.00{\rm{ s}}$ to the time it stopped is ${\rm{609 rad}}$ .