Laws Of Motion Question 403

Question: The member OA rotates about a horizontal axis through O with a constant counter clockwise velocity $ \omega =3 $

$ r/r $ . As it passes the position$ \theta =0^{o} $ , a small mass m is placed upon it at a radial distance r = 0.5 m. If the mass is observed to slip at $ \theta =37^{o} $ , the coefficient of friction between the mass & the member is:

Options:

A) $ \frac{3}{16} $

B) $ \frac{9}{16} $

C) $ \frac{4}{9} $

D) $ \frac{5}{9} $

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Answer:

Correct Answer: A

Solution:

As the mass is at the verge of slipping

$ \therefore $ $ mg\sin 37-\mu mg\cos 37=m{{\omega }^{2}}r $

$ 6-8\mu =4.5 $

$ \therefore $ $ \mu =\frac{3}{16} $



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