A uniformly dense solid cylinder of mass M and radius R is released from rest on an inclined plane and starts performing pure rolling. During its downward journey along the incline the cylinder moves distance l, the angle of inclination from horizontal is $α$ and the coefficient of friction is given as $μ .$ [The acceleration due to gravity is g downwards. Air resistance is not present.]

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

The final angular speed of cylinder is $\sqrt{\frac{2}{3} \frac{g l \sin α}{R^{2}}}$

The minimum coefficient of friction required so that there is no slipping is $\frac{\tan α}{3}$

The final angular speed of cylinder is $\sqrt{\frac{4}{3} \frac{g l \sin α}{R^{2}}}$

The acceleration of center of mass of cylinder is $\frac{2 g \sin α}{3}$

answer is A, B, C.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

For [A]:

$\begin{array}{l} a = \frac{g \sin α}{1 + \frac{K^{2}}{R^{2}}} = \frac{2}{3} g \sin α \\ F o r [B] : \\ V^{2} = 0 + 2 \times \frac{2}{3} g \sin α l \\ \Rightarrow V = \sqrt{\frac{4}{3}} g l \sin α \Rightarrow ω = \frac{V}{R} = \sqrt{\frac{4 g l \sin α}{3 R^{2}}} \\ F o r [C] : \end{array}$

For pure rolling

$\begin{array}{l} a = α R \Rightarrow \frac{M g \sin α - f}{M} = \frac{f \times R^{2}}{\frac{M R^{2}}{2}} \\ \Rightarrow M g \sin α - f = 2 f \Rightarrow f = \frac{M g \sin α}{3} \\ \Rightarrow F o r minimum μ, μ m g \cos α = \frac{M g \sin α}{3} \Rightarrow μ_{\min} = \frac{\tan α}{3} \end{array}$