(C) For linear motion:$F - f = Ma$ ---$(1)$For rotational motion about the center of mass:$\tau = I \alpha$$f \cdot R = (\frac{2}{5} M R^2) \cdot (\f

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(C) For linear motion:$F - f = Ma$ ---$(1)$For rotational motion about the center of mass:$\tau = I \alpha$$f \cdot R = (\frac{2}{5} M R^2) \cdot (\frac{a}{R})$$f = \frac{2}{5} Ma$ ---$(2)$From $(2)$,$Ma = \frac{5}{2} f$. Substituting this into $(1)$:$F - f = \frac{5}{2} f$$F = \frac{7}{2} f$For the sphere not to slip,the friction force $f$ must satisfy the condition $f \le \mu N$,where $N = Mg$.So,$f \le \mu Mg$.Substituting the maximum value of $f$ $(f_{max} = \mu Mg)$:$F_{max} = \frac{7}{2} \mu Mg$.

Class 11 Physics System of Particles and Rotational Motion Rolling motion on horizontal Surface

Difficult

$A$ horizontal force $F$ is applied at the center of a sphere as shown in the figure. The coefficient of friction between the sphere and the ground is $\mu$. What is the maximum value of $F$ if the sphere does not slip?

A
$\frac{5}{2} \mu Mg$
B
$\frac{4}{3} \mu Mg$
C
$\frac{7}{2} \mu Mg$
D
$\frac{9}{4} \mu Mg$

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System of Particles and Rotational Motion

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Rolling motion on horizontal Surface

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$A$ horizontal force $F$ is applied at the center of a sphere as shown in the figure. The coefficient of friction between the sphere and the ground is $\mu$. What is the maximum value of $F$ if the sphere does not slip?

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Consider a cylinder of mass $M$ resting on a rough horizontal rug that is pulled out from under it with acceleration $a$ perpendicular to the axis of the cylinder. What is the friction force $F_{friction}$ at point $P$? It is assumed that the cylinder does not slip.

$A$ disc rolls without slipping with a constant velocity. What fraction of its total kinetic energy is in the form of rotational kinetic energy?

Can the relation $v = r\omega$ be used for an object undergoing rolling motion with slipping? Why?

$A$ solid uniform sphere resting on a rough horizontal plane is given a horizontal impulse directed through its centre so that it starts sliding with an initial velocity $v_{0}$. When it finally starts rolling without slipping,the speed of its centre is

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