A solid sphere rolls purely on an inclined pl | vedclass

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(D) For a solid sphere rolling purely down an inclined plane:$(1)$ The friction force $f$ is given by $f = \frac{mg \sin 	heta}{1 + \frac{MR^2}{I}}$.

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$(3), (4)$

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(D) For a solid sphere rolling purely down an inclined plane:$(1)$ The friction force $f$ is given by $f = \frac{mg \sin 	heta}{1 + \frac{MR^2}{I}}$. For a solid sphere,$I = \frac{2}{5}MR^2$,so $f = \frac{mg \sin 	heta}{1 + 2.5} = \frac{2}{7}mg \sin 	heta$. Thus,statement $(1)$ is incorrect.$(2)$ In pure rolling,the point of contact is instantaneously at rest,so no work is done by friction. Therefore,it is not a dissipative force. Statement $(2)$ is incorrect.$(3)$ Friction acts up the incline,providing a torque that increases the angular velocity $\omega$ and a force that opposes the translational motion,thus decreasing the linear acceleration. Statement $(3)$ is correct.$(4)$ Since $f = \frac{2}{7}mg \sin 	heta$,if $	heta$ decreases,$\sin 	heta$ decreases,and thus the friction force $f$ decreases. Statement $(4)$ is correct.Therefore,statements $(3)$ and $(4)$ are correct.

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