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(D) Initially,the sphere rolls without slipping. The torque about the point of contact $(ICR)$ is $	au_{ICR} = FR = I_{ICR} \alpha_{ang}$,where $I_{I

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Vedclass · Answer

(D) Initially,the sphere rolls without slipping. The torque about the point of contact $(ICR)$ is $	au_{ICR} = FR = I_{ICR} \alpha_{ang}$,where $I_{ICR} = I_{cm} + mR^2 = \frac{2}{5}mR^2 + mR^2 = \frac{7}{5}mR^2$.Thus,$\alpha_{ang} = \frac{FR}{I_{ICR}} = \frac{(\alpha t)R}{\frac{7}{5}mR^2} = \frac{5 \alpha t}{7mR}$.The linear acceleration is $a = \alpha_{ang} R = \frac{5 \alpha t}{7m}$.Using $F - f = ma$,we get $f = F - ma = \alpha t - \frac{5 \alpha t}{7} = \frac{2 \alpha t}{7}$.Rolling continues until $f = \mu_s mg$,i.e.,$\frac{2 \alpha t}{7} = \mu_s mg$,which gives $t = \frac{7 \mu_s mg}{2 \alpha}$.For $t > \frac{7 \mu_s mg}{2 \alpha}$,the sphere slips,and kinetic friction $f_k = \mu_k mg$ acts.The new acceleration is $a' = \frac{F - f_k}{m} = \frac{\alpha t - \mu_k mg}{m} = \frac{\alpha t}{m} - \mu_k g$.Comparing the slopes: the initial slope is $\frac{5 \alpha}{7m}$ and the final slope is $\frac{\alpha}{m}$. Since $\frac{\alpha}{m} > \frac{5 \alpha}{7m}$,the slope increases after slipping. This corresponds to graph $D$.

Column-$I$	Column-$II$
$(1)$ Static friction	$(a)$ Limiting friction
$(2)$ Rolling friction	$(b)$ Ball bearing
	$(c)$ Object moving on a road

Similar Questions

Match Column-$I$ with Column-$II$ correctly.

Column-$I$ Column-$II$

$(1)$ Static friction $(a)$ Limiting friction

$(2)$ Rolling friction $(b)$ Ball bearing

$(c)$ Object moving on a road

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