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, [tex]\omega=2\ rev/sec=12.56\ rad/s[/tex]
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 :
[tex]f=mr\omega^2[/tex]
[tex]f=0.05\times 0.17\times (12.56)^2[/tex]
f = 1.34 N
So, the frictional force acting on the disk is 1.34 N. Hence, this is the required solution.
