Question

A block of mass $m$ is placed on a vertical fixed circular track and then it is given velocity $v$ along the track at position $A$ on track. The coefficient of friction between the block and the track varies with the angle $\theta$ . If the block moves on track with constant speed then the coefficient of friction is

• A. $\mu =\frac{sin\theta }{cot\theta +\frac{v^2}{Rg}}$
• B. $\mu =\frac{sin\theta }{cos\theta +\frac{v^2}{Rg}}$
• C. $\mu =\frac{tan\theta }{cos\theta +\frac{v^2}{Rg}}$
• D. $\mu =\frac{cos\theta }{tan\theta +\frac{v^2}{Rg}}$

Answer 1

Answer: (B)

Since the block is moving with a constant speed, at any instant its tangential acceleration is zero. Hence,
$f=mgsin\theta$
Also,
$N-mgcos\theta =\frac{m{v}^2}{R}$
$N=\frac{m{v}^2}{R}+mgcos\theta$
$f=\mu N=\mu \frac{m{v}^2}{R}+\mu mgcos\theta$
$mgsin\theta =\mu \frac{m{v}^2}{R}+\mu mgcos\theta$
$\Rightarrow \mu =\frac{sin\theta }{cos\theta +\frac{v^2}{Rg}}$