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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 $P$
Question

NTA AbhyasNTA Abhyas 2020

Solution:

$mgh=\frac{1}{2}mv^{2}+\frac{1}{2}I\omega ^{2}$
$7mgR=\frac{1}{2}mv^{2}+\frac{1}{2}\times \frac{2}{5}mv^{2}$
$mv^{2}=10mgR\Rightarrow v=\sqrt{10 gR}$