Question 142: Work, Energy and Power & Circular Motion

Question

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of $3.5$ revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s
2
)

Accepted Answer

We have
$m{\omega ^2}r = \mu mg$ ...... (1)
Given : rate of ro
tation = 3.5 rev/s
$\Rightarrow$ 1 revolution = 2$\pi$ rad
That is, 3.5 revolutions = 3.5 $	imes$ 2$\pi$ rad
Therefore, $\omega$ = 3.5 $	imes$ 2$\pi$ rad/s
r = 1.25 cm = 1.25 $	imes$ 10
$-$2
m
Thus, from Eq. (1), we have
$m{\omega ^2}r = \mu mg \Rightarrow {\omega ^2}r = \mu g$
$ \Rightarrow \mu  = {{{\omega ^2}r} \over g} = {{{{(3.5 	imes 2\pi )}^2}(1.25 	imes {{10}^{ - 2}})} \over {10}}$
$ \Rightarrow \mu  = 0.60$

Answer

0.5

Answer

0.3

Answer

0.7

Answer

0.6

Work, Energy and Power & Circular Motion

A disc rotates about its axis of symmetry in a horizontal plane at a steady rate of 3.5 revolutions per second. A coin placed at a distnce of 1.25 cm from the axis of rotation remains at rest on the disc. The coefficient of friction between the coin and the disc is : (g = 10 m/s 2 )

Choose the correct answer: