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(B) For a disc of mass $M$ and radius $R$ rolling down an inclined plane of angle $	heta$,the equations of motion are:$1$. Translational motion: $Mg

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$\frac{5}{3 \sqrt{2}} \, N$

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(B) For a disc of mass $M$ and radius $R$ rolling down an inclined plane of angle $	heta$,the equations of motion are:$1$. Translational motion: $Mg \sin 	heta - f = Ma_{cm}$$2$. Rotational motion about the center of mass: $	au = I_{cm} \alpha \implies fR = (\frac{1}{2} MR^2) \alpha$Since the condition for pure rolling is $a_{cm} = R\alpha$,we have $\alpha = \frac{a_{cm}}{R}$.Substituting this into the torque equation: $fR = \frac{1}{2} MR^2 (\frac{a_{cm}}{R}) \implies f = \frac{1}{2} Ma_{cm} \implies Ma_{cm} = 2f$.Substituting $Ma_{cm} = 2f$ into the translational equation: $Mg \sin 	heta - f = 2f \implies Mg \sin 	heta = 3f \implies f = \frac{Mg \sin 	heta}{3}$.Given: $M = 0.5 \, kg$,$	heta = 45^{\circ}$,$g = 10 \, m/s^2$.$f = \frac{0.5 	imes 10 	imes \sin(45^{\circ})}{3} = \frac{5 	imes (1/\sqrt{2})}{3} = \frac{5}{3 \sqrt{2}} \, N$.

$A$ disc of radius $20 \, cm$ and mass $0.5 \, kg$ is rolling on an inclined plane with an angle of inclination $45^{\circ}$. Find the friction force required for the disc to perform pure rolling.

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