A small ball of mass m and radius r=R/10 rolls without slipping along the track shown in the figure. The radius of circular part of the track is R. If…

Question

A small ball of mass m and radius r=R/10 rolls without slipping along the track shown in the figure. The radius of circular part of the track is R. If the ball starts from rest at a height of 8R above the bottom, the normal force on the ball at the point P is

MARKS Content Team · Accepted Answer

Correct answer: (B) 100/9mg. Using Work-Energy Theorem 7mgR=1/2mv^2+1/22/5mr^2×ω^2 since the sphere is rolling without slipping v=rω ∴7mgR=1/2× 7/5mr^2× v^2r^2 ⇒v^2=10gR at point P, we can consider sphere as a point object doing circular motion in a circle of the radius (R-r) with velocity v, Hence at point P N=mv^2R-r=10mgR/(R-R10)=10mgR/9R10 ( given r=R/10 ) or N=100/9mg

A small ball of mass $m$ and radius $r = \frac{R}{10}$ rolls without slipping along the track shown in the figure. The radius of circular part of the track is $R$ . If the ball starts from rest at a height of $8 R$ above the bottom, the normal force on the ball at the point $P$ is

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More Rotational Motion Questions

A small ball of mass m and radius r=R10 rolls without slipping along the track shown in the figure. The radius of circular part of the track is R. If the ball starts from rest at a height of 8R above the bottom, the normal force on the ball at the point P is

Solution

More Rotational Motion Questions

A small ball of mass $m$ and radius $r = \frac{R}{10}$ rolls without slipping along the track shown in the figure. The radius of circular part of the track is $R$ . If the ball starts from rest at a height of $8 R$ above the bottom, the normal force on the ball at the point $P$ is