A solid sphere and a hollow cylinder roll up | vedclass

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

Question

(B) Using the principle of conservation of energy,the loss in kinetic energy equals the gain in potential energy.$mgh = K.E._{total} = \frac{1}{2}mv^2

Vedclass · Accepted Answer

$7$

Vedclass · Answer

(B) Using the principle of conservation of energy,the loss in kinetic energy equals the gain in potential energy.$mgh = K.E._{total} = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$Since the object rolls without slipping,$\omega = \frac{v}{R}$ and $I = Mk^2$,where $k$ is the radius of gyration.$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$$h = \frac{v^2}{2g}(1 + \frac{k^2}{R^2})$Thus,$h \propto (1 + \frac{k^2}{R^2})$.For a solid sphere,$\frac{k^2}{R^2} = \frac{2}{5}$,so $h_1 \propto (1 + \frac{2}{5}) = \frac{7}{5}$.For a hollow cylinder,$\frac{k^2}{R^2} = 1$,so $h_2 \propto (1 + 1) = 2$.Therefore,$\frac{h_1}{h_2} = \frac{7/5}{2} = \frac{7}{10}$.Given $\frac{h_1}{h_2} = \frac{n}{10}$,we find $n = 7$.

Similar Questions

$A$ solid cylinder rolls down an inclined plane without slipping. Which of the following statements is correct?

$A$ cylinder rolls down two different inclined planes of the same height but of different inclinations.

$A$ spherical uniform body of radius $R$,mass $M$,and moment of inertia $I$ rolls down (without slipping) on an inclined plane making an angle $\theta$ with the horizontal. Then its acceleration is

$A$ cylinder and a ring of same mass $M$ and radius $R$ are placed on the top of a rough inclined plane of inclination $\theta$. Both are released simultaneously from the same height $h$. Choose the correct statement$(s)$ related to the motion of each body.

Three bodies: a ring,a solid cylinder,and a solid sphere,roll down an inclined plane without slipping. They start from rest. Which of the bodies reaches the bottom of the plane with the minimum velocity?

Vedclass Test Series

Exam Paper Generator

Online Exam Module