Question

A 50 g disk sits on a horizontally rotating turntable. The turntable makes exactly 2 revolutions each second. The disk is located 17 cm from the axis of rotation of the turntable. (a) What is the frictional force acting on the disk?

Answer 1

Answer:

Frictional force, f = 1.34 N

Explanation:

It is given that,

Mass of the disk, m = 50 g = 0.05 kg

The angular speed of the turntable, $\omega=2\ rev/sec=12.56\ rad/s$

The radius of the disk, r = 17 cm = 0.17 m

(a) Let f is the frictional force acting on the disk. The frictional force acting on the disk in rotational motion is given by :

$f=mr\omega^2$

$f=0.05\times 0.17\times (12.56)^2$

f = 1.34 N

So, the frictional force acting on the disk is 1.34 N. Hence, this is the required solution.