Question

A phonograph turntable rotating at 33 1/3 rev/min slows down and stops in 1.0 min. (a) What is the turntable’s angular acceleration, in radians/s², assuming it is constant? (b) How many revolutions does the turntable make while stopping?

a) (a) ( -0.058 , {rad/s}² ), (b) ( 16.7 , {rev} )
b) (a) ( -0.083 , {rad/s}² ), (b) ( 20 , {rev} )
c) (a) ( -0.058 , {rad/s}² ), (b) ( 20 , {rev} )
d) (a) ( -0.083 , {rad/s}² ), (b) ( 16.7 , {rev} )

Answer 1

The turntable's angular acceleration is -0.083 rad/s² and it makes 20 revolutions while stopping.

Step-by-step explanation:

To determine the turntable's angular acceleration, we need to use the equation:

angular acceleration = (final angular velocity - initial angular velocity) / time

Given that the turntable rotates at 33 1/3 rev/min, which is equivalent to 33 1/3 * 2π rad/min, and it slows down and stops in 1.0 min, we can calculate:

initial angular velocity = 33 1/3 * 2π rad/min

final angular velocity = 0 rad/min

angular acceleration = (0 - (33 1/3 * 2π)) / 1.0 = -0.083 rad/s²

To calculate the number of revolutions the turntable makes while stopping, we can use the equation:

number of revolutions = (final angular velocity - initial angular velocity) / (2π)

Given that the initial angular velocity is 33 1/3 * 2π rad/min and the final angular velocity is 0 rad/min, we can calculate:

number of revolutions = (0 - (33 1/3 * 2π)) / (2π) = 20 rev