A body slides down a smooth inclined plane of | vedclass

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

Question

(A) For a body sliding down a smooth inclined plane, there is no friction, so all potential energy is converted into translational kinetic energy: $mg

Vedclass · Accepted Answer

$\frac{V}{\sqrt{2}}$

Vedclass · Answer

(A) For a body sliding down a smooth inclined plane, there is no friction, so all potential energy is converted into translational kinetic energy: $mgh = \frac{1}{2}mV^2$, which gives $V = \sqrt{2gh}$ or $V^2 = 2gh$.For a ring rolling down the same inclined plane, the potential energy is converted into both translational and rotational kinetic energy: $mgh = \frac{1}{2}mv'^2 + \frac{1}{2}I\omega^2$.Since $I = mr^2$ and $\omega = v'/r$ for a ring, we have $mgh = \frac{1}{2}mv'^2 + \frac{1}{2}(mr^2)(v'/r)^2 = \frac{1}{2}mv'^2 + \frac{1}{2}mv'^2 = mv'^2$.Thus, $v'^2 = gh$.Substituting $gh = V^2/2$ into the equation, we get $v'^2 = V^2/2$, which implies $v' = \frac{V}{\sqrt{2}}$.

$A$ body slides down a smooth inclined plane of inclination $\theta$ and reaches the bottom with velocity $V$. If the same body is a ring which rolls down the same inclined plane, then the linear velocity at the bottom of the plane is:

Similar Questions

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 ........

$A$ solid sphere is rolling on a frictionless surface. It moves with a translational velocity $v \ m/s$ and climbs an inclined plane as shown in the figure. What should be the minimum value of $v$?

$A$ solid sphere of mass $4 \ kg$ and radius $28 \ cm$ is on an inclined plane. If the acceleration of the sphere when it rolls down without sliding is $3.5 \ m \ s^{-2}$,then the acceleration of the sphere when it slides down without rolling is (in $m \ s^{-2}$)

$A$ uniform solid cylinder of mass $m$ and radius $r$ rolls along an inclined rough plane of inclination $45^{\circ}$. If it starts to roll from rest from the top of the plane,then the linear acceleration of the cylinder axis will be:

$A$ ring,sphere,and disc are rolling down an inclined plane from the same height. Find the wrong statement: (where $t$ is the time of descent,$a$ is the acceleration,and $v$ is the speed at the bottom).

Vedclass Test Series

Exam Paper Generator

Online Exam Module