Question

A rectangular rigid fixed block has a long horizontal edge. A solid homogeneous cylinder of radius $r$ is placed horizontally at rest with its length parallel to the edge such that the axis of the cylinder and the edge of the block are in the same vertical plane. There is sufficient friction present at the edge so that a very small displacement causes the cylinder to roll off the edge without slipping. Determine :

(a) The angle $θ_{c}$ through which the cylinder rotates before it leaves contact with the edge.

(b) The speed of the centre of mass of the cylinder before leaving contact with the edge.

(c) The ratio of translational to rotational kinetic energies of the cylinder when its centre of mass is in horizontal line with the edge.

Accepted Answer

From FBD of cylinder

$m g \cos (θ) - N = \frac{m v^{2}}{r}$

At the time whne it loses the contact with edge N=0

$m g \cos (θ) = \frac{m v^{2}}{r} \Rightarrow v^{2} = r g \cos (θ) . . . . . . . (1)$

Its loss in PE will be equal to gain in RKE because centre of mass does not translate

$m g r [1 - \cos (θ)] = \frac{1}{2} I ω^{2} I a b o u t e d g e = \frac{3}{2} m r^{2} \Rightarrow m g r [1 - \cos (θ)] = \frac{1}{2} \times \frac{3}{2} m r^{2} \times \frac{v^{2}}{r^{2}} a t n o s l i p p i n g v = r ω v^{2} = \frac{4}{3} g r [1 - \cos (θ)]$

From equation (1)

$r g \cos (θ) = \frac{4}{3} g r [1 - \cos (θ) \Rightarrow \cos (θ) = \frac{4}{7} \Rightarrow θ = \cos^{- 1} (\frac{4}{7})$ .

(b).Speed of centre of mass,

$v^{2} = r g \cos (θ) \Rightarrow v^{2} = r g \times \frac{4}{7} \Rightarrow v_{c m} = \sqrt{\frac{4 r g}{7}}$

(c). $A t t h e t i e o f l e a v i n g c o n t a c t v^{2} = \frac{4}{7} g r ω^{2} = \frac{v^{2}}{r^{2}} = \frac{\frac{4}{7} g r}{r^{2}} \Rightarrow ω^{2} = \frac{4 g}{7 r} K E_{R} = \frac{1}{2} I ω^{2} = \frac{1}{2} \times \frac{1}{2} m r^{2} \times \frac{4 g}{7 r} K E_{R} = \frac{m g r}{7} K E_{T} = m g r - \frac{m g r}{7} K E_{T} = \frac{6 m g r}{7} \Rightarrow \frac{K E_{T}}{K E_{R}} = 6$

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