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Chapter 5: Problem 49

A turntable reaches an angular speed of \(33.3 \mathrm{rpm}\) in $2.0 \mathrm{s}$ starting from rest. (a) Assuming the angular acceleration is constant, what is its magnitude? (b) How many revolutions does the turntable make during this time interval?

Short Answer

Expert verified
Answer: The magnitude of the angular acceleration is 1.74 rad/s², and the turntable makes about 0.554 revolutions during the 2.0-second time interval.

Step by step solution

01

Convert to radians

First, we should convert the given angular velocity from rpm (revolutions per minute) to radians per second. $$ \omega_f = 33.3 \frac{rev}{min} \times \frac{2 \pi \ rad}{rev} \times \frac{1 \ min}{60 \ s} \approx 3.48 \; \frac{rad}{s} $$
02

Determine the angular acceleration

Now, we will use the linear acceleration formula: $$ \omega_f = \omega_i + \alpha t $$ Since the turntable starts from rest, the initial angular velocity (\(\omega_i\)) is 0. Given the final angular velocity \(\omega_f = 3.48 \; \frac{rad}{s}\) and the time interval \(t = 2.0 \ s\), we can solve for the angular acceleration \(\alpha\): $$ \alpha = \frac{\omega_f - \omega_i}{t} = \frac{3.48 \; \frac{rad}{s}}{2.0 \, s} = 1.74 \; \frac{rad}{s^2} $$ So, the magnitude of the angular acceleration is 1.74 rad/s².
03

Calculate the angular displacement

To find the number of revolutions made by the turntable during the 2.0-second time interval, we need to calculate the angular displacement \(\theta\). We can use the following formula for angular displacement: $$ \theta = \omega_i t + \frac{1}{2} \alpha t^2 $$ Since \(\omega_i = 0\), the equation simplifies to: $$ \theta = \frac{1}{2} \alpha t^2 $$ Plugging in the values for \(\alpha\) and \(t\), we have: $$ \theta = \frac{1}{2} (1.74 \; \frac{rad}{s^2})(2.0 \, s)^2 \approx 3.48 \, rad $$
04

Determine the number of revolutions

Finally, we need to convert the angular displacement in radians to the number of revolutions. We can do this by dividing the angular displacement by \(2\pi\) since there are \(2 \pi\) radians in one revolution: $$ \text{Number of revolutions} = \frac{3.48 \, rad}{2 \pi \, rad/rev} \approx 0.554 \, rev $$ So, the turntable rotates approximately 0.554 revolutions during the 2.0-second time interval. In conclusion, the magnitude of the angular acceleration is 1.74 rad/s², and the turntable makes about 0.554 revolutions during the 2.0-second time interval.

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