A solid cylinder is placed on a rough surface | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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

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

Vedclass · Answer

(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:

Similar Questions

$A$ body rolls down an inclined plane without slipping. The kinetic energy of rotation is $50 \%$ of its translational kinetic energy. The body is :

$A$ solid cylinder and a disc of same radii are allowed to roll down a rough inclined plane from the top of a plane. The ratio of their times taken to reach the bottom of the inclined plane is

$A$ small object of uniform density rolls up a curved surface with an initial velocity $v$. It reaches up to a maximum height of $\frac{3 v^2}{4 g}$ with respect to the initial position. The object is

$A$ spherical ball of mass $20 \ kg$ is stationary at the top of a hill of height $100 \ m$. It rolls down a surface to the ground,then climbs up another hill of height $30 \ m$ and finally rolls down to a horizontal base at a height of $20 \ m$ above the ground. The velocity attained by the ball is

Two discs having masses in the ratio $1:2$ and radii in the ratio $1:8$ roll down without slipping one by one from an inclined plane of height $h$. The ratio of their linear velocities on reaching the ground is ........

Vedclass Test Series

Exam Paper Generator

Online Exam Module