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: