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alexdok [17]
3 years ago
13

An electric motor rotating a workshop grinding wheel at a rate of 1.31 ✕ 102 rev/min is switched off. Assume the wheel has a con

stant negative angular acceleration of magnitude 3.40 rad/s2. (a) How long does it take for the grinding wheel to stop? (b) Through how many radians has the wheel turned during the interval found in (a)?
Physics
1 answer:
antoniya [11.8K]3 years ago
6 0

(a) 4.03 s

The initial angular velocity of the wheel is

\omega_i = 1.31 \cdot 10^2 \frac{rev}{min} \cdot \frac{2\pi rad/rev}{60 s/min}=13.7 rad/s

The angular acceleration of the wheel is

\alpha = -3.40 rad/s^2

negative since it is a deceleration.

The angular acceleration can be also written as

\alpha = \frac{\omega_f - \omega_i}{t}

where

\omega_f = 0 is the final angular velocity (the wheel comes to a stop)

t is the time it takes for the wheel to stop

Solving for t, we find

t=\frac{\omega_f - \omega_i }{\alpha}=\frac{0-13.7 rad/s}{-3.40 rad/s^2}=4.03 s

(b) 27.6 rad

The angular displacement of the wheel in angular accelerated motion is given by

\theta= \omega_i t + \frac{1}{2}\alpha t^2

where we have

\omega_i=13.7 rad/s is the initial angular velocity

\alpha = -3.40 rad/s^2 is the angular acceleration

t = 4.03 s is the total time of the motion

Substituting numbers, we find

\theta= (13.7 rad/s)(4.03 s) + \frac{1}{2}(-3.40 rad/s^2)(4.03 s)^2=27.6 rad

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