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Chapter 8: Problem 12

A turntable rotating at 78 rev/min slows down and stops in \(32 \mathrm{~s}\) after the motor is turned off. \((a)\) Find its (constant) angular acceleration in rev \(/ \mathrm{min}^{2} .(b)\) How many revolutions does it make in this time?

Short Answer

Expert verified
The angular deceleration is -146.17 rev/min^2 and the turntable makes approximately 42 revolutions before coming to a stop.

Step by step solution

01

Convert Units

Firstly, convert the time from seconds to minutes, so it is the same unit as given in rev/min. Thus, the time will be 32/60 = 0.533 min.
02

Find Angular Acceleration

Use the formula for acceleration \( \alpha = \frac{\Delta \omega}{\Delta t} = \frac{\omega_{f} - \omega_{i}}{\Delta t} \). From the problem, we know that \( \omega_{i} = 78 \ rev/min \) and \( \omega_{f} = 0 \ rev/min \). Substitute the values and we have \( \alpha = \frac{0 - 78}{0.533} = -146.17 \ rev/min^2 \). The acceleration is negative, indicating a slowdown.
03

Find Total Number of Revolutions

Use the displacement formula \( \Delta \theta = \omega_{i} \Delta t+0.5 \alpha \Delta t^{2} \). Substituting the values we have \( \Delta \theta = 78 * 0.533 + 0.5*(-146.17) * (0.533)^2 = 41.51 \ rev \). The negative term in the formula comes from the slowing down of the turntable. Hence, the turntable makes approximately 42 revolutions before coming to a stop.

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Most popular questions from this chapter

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