A solid ball of mass m and radius r rolls without slipping along the track shown in the fig. The radius of the circular part of the track is R. The ball starts rolling down the track from rest from a height of 8R from the ground level.What will be the velocity when the ball reaches the point P

Question

A solid ball of mass m and radius r rolls without slipping along the track shown in the fig. The radius of the circular part of the track is R. The ball starts rolling down the track from rest from a height of 8R from the ground level.What will be the velocity when the ball reaches the point P <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-sqwmpyl6you5li22.jpg" /> (A) √g R (B) √5 g R (C) √10 g R (D) √3 g R

Tardigrade Admin · Accepted Answer

mgh=(1/2)mv2+(1/2)Iω 2 7mgR=(1/2)mv2+(1/2)× (2/5)mv2 mv2=10mgR⇒ v=√10 gR

$g R$

$5 g R$

$10 g R$

$3 g R$

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$7 m g R = \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{2}{5} m v^{2}$
$m v^{2} = 10 m g R \Rightarrow v = 10 g R$

Q. A solid ball of mass m and radius r rolls without slipping along the track shown in the fig. The radius of the circular part of the track is R. The ball starts rolling down the track from rest from a height of 8R from the ground level.What will be the velocity when the ball reaches the point P

gR​

5gR​

10gR​

3gR​

mgh=21​mv2+21​Iω2 7mgR=21​mv2+21​×52​mv2 mv2=10mgR⇒v=10gR​

$g R$

$5 g R$

$10 g R$

$3 g R$

$m g h = \frac{1}{2} m v^{2} + \frac{1}{2} I ω^{2}$
$7 m g R = \frac{1}{2} m v^{2} + \frac{1}{2} \times \frac{2}{5} m v^{2}$
$m v^{2} = 10 m g R \Rightarrow v = 10 g R$