JEE MAIN - Physics (2018 - 15th April Evening Slot - No. 20)
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/s2)
0.5
0.3
0.7
0.6
Explanation
We have
$$m{\omega ^2}r = \mu mg$$ ...... (1)
Given : rate of rotation = 3.5 rev/s
$$\Rightarrow$$ 1 revolution = 2$$\pi$$ rad
_15th_April_Evening_Slot_en_20_1.png)
That is, 3.5 revolutions = 3.5 $$\times$$ 2$$\pi$$ rad
Therefore, $$\omega$$ = 3.5 $$\times$$ 2$$\pi$$ rad/s
r = 1.25 cm = 1.25 $$\times$$ 10$$-$$2 m
Thus, from Eq. (1), we have
$$m{\omega ^2}r = \mu mg \Rightarrow {\omega ^2}r = \mu g$$
$$ \Rightarrow \mu = {{{\omega ^2}r} \over g} = {{{{(3.5 \times 2\pi )}^2}(1.25 \times {{10}^{ - 2}})} \over {10}}$$
$$ \Rightarrow \mu = 0.60$$
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