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A uniform spherical shell of mass 'm' and radius 'r' is placed rotating such that its center of mass remains at rest in two different scenarios as shown in figure (A) and (B). If the coefficient of friction between all the surfaces is '\mu' then the ratio of the frictional force exerted by the ground in situations (A) and (B) will be-

(A)(B)

 

Option: 1

\frac{1}{1-\mu}


Option: 2

\frac{1}{1+\mu}


Option: 3

\frac{1}{1-\mu^{2}}


Option: 4

\frac{1}{1+\mu^{2}}


Answers (1)

best_answer

In figure A -

\begin{aligned} N_{1}+\mu N_{2} & =m g \\ \mu N_{1} & =N_{2} \end{aligned}

\begin{aligned} & N_{1}=\frac{m g}{\left(1+\mu^{2}\right)} \\ & f_{1}=\mu N_{1}=\frac{\mu m g}{\left(1+\mu^{2}\right)} \end{aligned}

In figure B-

\begin{aligned} & N_{1}^{\prime}=f_{2}^{\prime}+m g \\ & N_{1}^{\prime}=\mu N_{2}^{\prime}+m g \end{aligned}

As, \begin{aligned} & N_{2}^{\prime}=0 \\ & N_{1}^{\prime}=m g \\ & f_{1}^{\prime}=\mu N_{1}^{\prime}=\mu m g \end{aligned}

\text { so, } \frac{f_{1}}{f_{1}^{\prime}}=\frac{\mu \operatorname{mg} g}{\left(1+\mu^{2}\right)} \times \frac{1}{\mu \operatorname{mg} g}=\frac{1}{1+\mu^{2}}

Posted by

Kuldeep Maurya

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