Question

An antique carousel that’s powered by a large electric motor undergoes constant angular acceleration from rest to full rotational speed in 5.00 seconds. When the ride ends, a brake causes it to decelerate steadily from full rotational speed to rest in 11.0 seconds. Compare the torque that starts the carousel to the torque that stops it. τstart/τstop = the tolerance is +/-2%

Answer 1

1) For a rotating object, we can write the analogous of Newton's second law as
$\tau = I \alpha$
where $\tau$ is the torque applied to the object, I is the moment of inertia and $\alpha$ is the angular acceleration.

2) We can write the ratio between the starting torque and the stopping torque as:
$\frac{\tau _{start}}{\tau _{stop}} = \frac{I \alpha _{start}}{I \alpha _{stop}} = \frac{ \alpha _{start}}{\alpha _{stop}}$

3) Then, we can rewrite the two angular accelerations. The general formula is
$\alpha = \frac{\omega _f - \omega _i}{t}$
where $\omega _f$ and $\omega _i$ are the final and initial angular velocities, while t is the time. 

4) Let's start by calculating the starting acceleration. In this case, the initial velocity is zero, while the final velocity is the full rotational speed (let's call it $\omega _{max}$, and the time is t=5.0 s:
$\alpha _{start} = \frac{\omega_{max}-0}{5 s}= \frac{\omega_{max}}{5 s}=$
When the carousel stops, instead, the initial velocity is $\omega _{max}$ while the final velocity is 0, and the time is t=11.0 s:
$\alpha _{stop} = \frac{0-\omega _{max}}{11 s}= \frac{-\omega _{max}}{11 s}$
Therefore, if we replace the two angular accelerations that we found into the formula of the torque ration written at step 2), we find:
$\frac{\tau _{start}}{\tau _{stop}} = - \frac{11}{5}$
where the negative sign simply means that the two torques have opposite direction. Therefore, the ratio between starting and stopping torque is 11:5.