A rigid body of mass M and radius R rolls wit | vedclass

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(C) For a rigid body rolling without slipping on an inclined plane, the force of friction $f$ is given by the formula: $f = \frac{Mg \sin 	heta}{1 +

Vedclass · Accepted Answer

$A-II, B-III, C-IV, D-II$

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(C) For a rigid body rolling without slipping on an inclined plane, the force of friction $f$ is given by the formula: $f = \frac{Mg \sin 	heta}{1 + \frac{MR^2}{I}}$.$(A)$ For a ring, $I = MR^2$. Thus, $f = \frac{Mg \sin 	heta}{1 + 1} = \frac{Mg \sin 	heta}{2}$.$(B)$ For a solid sphere, $I = \frac{2}{5}MR^2$. Thus, $f = \frac{Mg \sin 	heta}{1 + 2.5} = \frac{Mg \sin 	heta}{3.5}$.$(C)$ For a solid cylinder, $I = \frac{1}{2}MR^2$. Thus, $f = \frac{Mg \sin 	heta}{1 + 2} = \frac{Mg \sin 	heta}{3}$.$(D)$ For a hollow cylinder, $I = MR^2$. Thus, $f = \frac{Mg \sin 	heta}{1 + 1} = \frac{Mg \sin 	heta}{2}$.Matching the results: $A-II, B-I, C-III, D-II$.

Column-$I$	Column-$II$
$(A)$ Ring	$(I)$ $\frac{Mg \sin \theta}{3.5}$
$(B)$ Solid sphere	$(II)$ $\frac{Mg \sin \theta}{2}$
$(C)$ Solid cylinder	$(III)$ $\frac{Mg \sin \theta}{3}$
$(D)$ Hollow cylinder	$(IV)$ $\frac{Mg \sin \theta}{2.5}$

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