# A DVD rotates from rest to an angular speed of 31.4 rad/s in a time of 0.892 seconds. A) What is the angular acceleration of the disc (assuming uniform acceleration"). B) Through what angle does...

A DVD rotates from rest to an angular speed of 31.4 rad/s in a time of 0.892 seconds.

A) What is the angular acceleration of the disc (assuming uniform acceleration"). B) Through what angle does the disc turn while coming up to speed? C) How many revolutions does it go through? D) If the disc is 9.9 cm across at its widest point, find the tangential speed of a microbe riding on the rim of the disc when t=0.892s. E) What is the magnitude of the tangential acerleration of the microbe at the given time?

### 1 Answer | Add Yours

The uniform angular acceleration is (like the linear acceleration)

`epsilon = (Delta(omega))/(Delta(t)) =31.4/0.892 =35.2 (rad)/s^2`

The angle traveled in time `t` is (like the linear space)

`alpha = epsilon*t^2/2 =35.2*0.892^2/2 =14.0044 rad`

To find the number of revolutions made we know that one complete revolution is through the angle `2*pi`

`N = alpha/(2*pi) =14.0044/(2*pi)= 2.23 revolutions`

At `t=0.892 seconds` the angular speed is `omega=31.4 (rad)/s` . The linear speed at a distance `R` from center is just

`V= omega*R =31.4*0.099 =3.1086 m/s`

The tangential acceleration at a distance `R` from center is (like the tangential speed)

`a= epsilon*R =35.2*0.099 =3.4848 m/s^2`

**Sources:**