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(A) For a body rolling down an inclined plane without slipping,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.For a soli

Vedclass · Accepted Answer

$\frac{1}{3} 	an 	heta$

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(A) For a body rolling down an inclined plane without slipping,the acceleration is given by $a = \frac{g \sin 	heta}{1 + \frac{k^2}{R^2}}$.For a solid cylinder,the radius of gyration $k$ is such that $k^2 = \frac{1}{2} R^2$,so $\frac{k^2}{R^2} = \frac{1}{2}$.Thus,$a = \frac{g \sin 	heta}{1 + 1/2} = \frac{2}{3} g \sin 	heta$.The force equation along the incline is $mg \sin 	heta - f = ma$,where $f$ is the friction force.Substituting $a$,we get $f = mg \sin 	heta - m(\frac{2}{3} g \sin 	heta) = \frac{1}{3} mg \sin 	heta$.Also,the torque equation about the center is $f R = I \alpha = (\frac{1}{2} m R^2)(\frac{a}{R}) = \frac{1}{2} m R a$.Substituting $a$,$f = \frac{1}{2} m (\frac{2}{3} g \sin 	heta) = \frac{1}{3} mg \sin 	heta$.Since $f \le \mu N$ and $N = mg \cos 	heta$,we have $\frac{1}{3} mg \sin 	heta \le \mu mg \cos 	heta$.Therefore,$\mu \ge \frac{1}{3} 	an 	heta$.

$A$ solid cylinder is placed on a rough surface of inclination $\theta$. The minimum value of the coefficient of static friction between the cylinder and the surface so that the cylinder rolls without slipping is:

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