A sphere is placed rotating with its centre initially at rest ina corner as shown in figure (a) &amp

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6.System of Particles and Rotational Motion

normal

$A$ sphere is placed rotating with its centre initially at rest ina corner as shown in figure $(a)$ & $(b)$. Coefficient of friction between all surfaces and the sphere is $\frac{1}{3}$. Find the ratio of the frictional force $\frac{{{f_a}}}{{{f_b}}}$ by ground in situations $(a)$ & $(b)$.

$1$

$\frac{9}{{10}}$

$\frac{{10}}{9}$

none

Solution

From the $FBD$ of figure a, we get the translational equilibrium in vertical direction,

$\mu N_{1}+N_{2}=m g…(i)$

In horizontal directions,

$N_{1}=\mu N_{2}$ $…(ii)$

Solving, Eqs.$(i)$ and $(ii),$ $N_{2}=\frac{m g}{1+\mu^{2}}$

The required friction is

$\mu N_{2}=f_{a}=\frac{3}{10} m g$

Now in case of figure $b,$ we see that the sphere has the tendency to move towards the right and thus there would be no interaction between the sphere and the vertical surface. Thus the normal reaction in the horizontal direction would be zero. Thus we get

$N_{1}=0 ; N_{2}=m g$

$f_{b}=\mu N_{2}=\frac{m g}{3}$

which gives $\frac{f_{a}}{f_{b}}=\frac{9}{10}$

Standard 11

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Solution

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