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dexar [7]
3 years ago
8

An antique carousel that’s powered by a large electric motor undergoes constant angular acceleration from rest to full rotationa

l 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%
Physics
1 answer:
SVEN [57.7K]3 years ago
8 0
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.

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