A rigid body of mass M and radius R rolls without slipping on an inclined plane of inclination θ, under gravity. Match the type of body Column I with magnitude of the force of friction Column II. Column I Column II A)

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Quantrex Academy · Accepted Answer

Correct answer: A. A B C D IV III II IV

if a is the acceleration of the center of mass of the rigid body and f the force of friction between the sphere and the plane, the equation of translatory and rotatory motion of the rigid body will be, Mgsinθ - f = Ma ⇒ f = M( ) + f and the relation between the moment of inertia and friction force, f × R = Iα ⇒ α = (f × R)/(I) , ⇒ Mgsinθ = M( f × R ^ 2 I ) + f ⇒ f = Mgsinθ M R ^ 2 I + 1 The value of the moment of inertia of solid sphere, I = (2)/(5)M R ^ 2 , for the ring, I = M R ^ 2 , for cylinder, I = (1)/(2) × M R ^ 2 and for hollow cylinder, I = M R ^ 2 it means, ⇒ f _ ring = (Mgsinθ)/(2), f _ hc = (Mgsinθ)/(2), f _ sc = (Mgsinθ)/(3) and f _ ring = (Mgsinθ)/(3 . 5),

Column I		Column II
A)	For ring	I)	$\dfrac{Mgsin\theta}{2 . 5}$
B)	For solid sphere	II)	$\dfrac{Mgsin\theta}{3}$
C)	For solid cylinder	III)	$\dfrac{Mgsin\theta}{3 . 5}$
D)	For hallow cylinder	IV)	$\dfrac{Mgsin\theta}{2}$

A rigid body of mass M and radius R rolls without slipping on an...

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