Question

A turntable that spins at a constant 74.0 rpm takes 3.10 s to reach this angular speed after it is turned on. Find its angular acceleration (in rad/s2), assuming it to be constant, and the number of degrees it turns through while speeding up.

Answer 1

Answer:

$2.5 rad/s^2, 688^{\circ}$

Explanation:

The angular acceleration is given by:

$\alpha = \frac{\omega_f - \omega_i}{t}$

where

$\omega_f = 74.0 rev/min \cdot (\frac{2\pi rad/rev}{60 s/min})=7.75 rad/s$ is the final angular speed

$\omega_i = 0$ is the initial angular speed

t = 3.10 s is the time interval

Solving the equation,

$\alpha = \frac{7.75 rad/s - 0}{3.10 s}=2.5 rad/s^2$

Now we can find the angular displacement by using:

$\theta = \omega_i t + \frac{1}{2}\alpha t^2$

Substituting,

$\theta=0+\frac{1}{2}(2.5 rad/s^2)(3.10 s)^2=12.0 rad$

In degrees:

$\theta = \frac{12.0 rad}{2\pi}\cdot 360^{\circ}=688^{\circ}$