Question

The old-fashioned record player had two speeds: one for long-playing albums (LP) and one for singles. The turntable speed for LPs was 33 1/3 rpm (revolutions per minute). This was equivalent to 100 revolutions every three minutes. The speed of a single was 45 rpm. Suppose it took 4.0 seconds for the turntable to ‘speed up’ from 33 1/3 rpm to 45 rpm when a switch was pressed to change the speed. What would be this angular acceleration in rad/sec2?

Answer 1

`\alpha = 0.305 rad/s^2`

Step-by-step explanation:

initial frequency of revolution is given as

`f_1 = 100/3 rpm`

now initial angular speed is

`\omega_i = 2\pi f`

`\omega_i = 2\pi((100)/(3* 60))`

`\omega_i = 3.49 rad/s`

Similarly final angular speed is given as

`\omega_f = 2\pi f_2`

`\omega_f = 2\pi((45)/(60))`

`\omega_f = 4.71 rad/s`

Now angular acceleration is given as

`\alpha = (\omega_f - \omega_i)/(\Delta t)`

`\alpha = (4.71 - 3.49)/(4)`

`\alpha = 0.305 rad/s^2`