A solid cylinder (i) rolls down and (ii) slid | vedclass

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

Question

(B) For a solid cylinder sliding down an inclined plane without friction,the acceleration is $a_{slide} = g \sin 	heta$.For a solid cylinder rolling

Vedclass · Accepted Answer

$2:3$

Vedclass · Answer

(B) For a solid cylinder sliding down an inclined plane without friction,the acceleration is $a_{slide} = g \sin 	heta$.For a solid cylinder rolling down an inclined plane without slipping,the acceleration is $a_{roll} = \frac{g \sin 	heta}{1 + \frac{I}{MR^2}}$.For a solid cylinder,the moment of inertia about its central axis is $I = \frac{1}{2} MR^2$.Substituting this into the rolling acceleration formula: $a_{roll} = \frac{g \sin 	heta}{1 + \frac{1/2 MR^2}{MR^2}} = \frac{g \sin 	heta}{1 + 0.5} = \frac{g \sin 	heta}{1.5} = \frac{2}{3} g \sin 	heta$.The ratio of the acceleration of rolling to sliding is $\frac{a_{roll}}{a_{slide}} = \frac{\frac{2}{3} g \sin 	heta}{g \sin 	heta} = \frac{2}{3}$.Thus,the ratio of the accelerations is $2:3$.

$A$ solid cylinder $(i)$ rolls down and $(ii)$ slides down an inclined plane. What is the ratio of the accelerations in these conditions?

Similar Questions

Three bodies,a ring,a solid cylinder,and a solid sphere,roll down the same inclined plane without slipping. They start from rest. The radii of the bodies are identical. Which of the bodies reaches the ground with maximum velocity?

$A$ cylinder of mass $m$ and radius $R$ rolls without slipping down an inclined plane of length $L$ and height $h$. What will be the velocity of its center of mass when the cylinder reaches the bottom?

$A$ disc of mass $M$ and radius $R$ rolls on a horizontal surface and then rolls up an inclined plane as shown in the figure. If the velocity of the disc is $v,$ the height to which the disc will rise will be

$A$ disc is given an angular velocity $\omega_0$ as shown in the figure and is kept on an inclined plane of inclination $\theta$ having a coefficient of friction $\mu = \tan \theta$. If the initial acceleration of the centre of mass is $a$ and the angular acceleration is $\alpha$,then:

The moment of inertia of a solid cylinder about its axis is $I$. It is allowed to roll down an incline without slipping. If its angular velocity at the bottom is $\omega$,then the total kinetic energy $(K.E.)$ of the cylinder will be:

Vedclass Test Series

Exam Paper Generator

Online Exam Module