Question

At time t=0 a grinding wheel has an angular velocity of 30.0 rad/s . It has a constant angular acceleration of 35.0 rad/s2 until a circuit breaker trips at time t = 2.00 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? Express your answer in radians.
B) At what time does the wheel stop? Express your answer in seconds.
C) What was the wheel's angular acceleration as it slowed down? Express your answer in radians per second per second.

Answer 1

Answer:

(A) 570 rad

(B) 10 s

(C) 12.5 rad/s²

Explanation:

The equations of motion for circular motions are used.

• Initial angular velocity,  $\omega_0 = 30.0 \text{ rad/s}$

• Angular acceleration, $\alpha =35.0 \text{ rad/s}^2$

(A)

At t = 2.00 s, the angular displacement, θ, is given by

$\theta = \omega_0t+\frac{1}{2}\alpha t^2 = (30\times 2) + \frac{1}{2}\times35\times2^2=60+70 = 130\text{ rad}$

After this time, it decelerates through an angular displacement of 440 rad.

Total angular displacement = 130 + 440 rad = 570 rad

(B)

At the time the circuit breaker tips, the angular velocity is given by

$\omega = \omega_0+\alpha t = 30.0+(35.0\times 2) = 30.0+70.0 =100.0\ \text{rad/s}$

This becomes the initial angular velocity for the decelerating motion. Because it stops, the final angular velocity is 0 rad/s. The time for this part of the motion is calculated thus:

$\theta_2 = \left(\dfrac{\omega_i+\omega_f}{2}\right)t$

Here, $\theta_2=440$ (the angular displacement during deceleration)

The subscripts, i and f, on ω denote the initial and final angular velocities during deceleration.

$\omega_i = 100$

$\omega_f = 0$

$t = \dfrac{2\theta_2}{\omega_i} = \dfrac{2\times400}{100} = 8\ \text{s}$

This is the time for deceleration. The deceleration began at t = 2 s.

Hence, the wheel stops at t = 2 + 8 = 10 s.

(C)

The deceleration is given by

$\alpha_R = \dfrac{\omega_f-\omega_i}{t} = \dfrac{0-100}{8} = -12.5\text{ rad/s}^2$

The negative sign appears because it is a deceleration.