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:

`(\tau _(start))/(\tau _(stop)) = (I \alpha _(start))/(I \alpha _(stop)) = ( \alpha _(start))/(\alpha _(stop))`

3) Then, we can rewrite the two angular accelerations. The general formula is

`\alpha = (\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) = (\omega_(max)-0)/(5 s)= (\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) = (0-\omega _(max))/(11 s)= (-\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:

`(\tau _(start))/(\tau _(stop)) = - (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.