Question

A non uniform ball of mass M and radius R, rolls from rest down a ramp and onto a circular loop of
radius r. The ball is initially at height h above the bottom of loop. At the bottom of loop the normal
force on ball is twice its weight. Moment of inertia of ball is given by 1 = BMR'. Find ß
(Assume h >> R and r >> R)
(A)h/R
(B) 2h/r
(C) 2h/r-1
(1) 2h/r +1​

Answer 1

Answer:

First of all , let's apply Conservation pf Mechanical Energy :

The whole Potential Energy of the ball at the starting point will be converted to Rolling Kinetic energy at the bottom of the loop .

$\boxed{ \sf{Mgh = \frac{1}{2}M {v}^{2} \bigg\{1 + \frac{ {k}^{2} }{ {R}^{2} } \bigg\}}}$

We also know that Radius of Gyration is denoted as k

$\boxed{ \sf{\beta M {R}^{2} = M {k}^{2} }}$

$\sf{\implies \: \beta = \bigg \{ \dfrac{ {k}^{2} }{ {R}^{2} } \bigg \}} \: ......(1)$

Now, let's draw FBD of ball at the bottom of loop.

$\sf{\therefore N - Mg = \dfrac{M {v}^{2} }{r} }$

$\sf{ \implies \: 2Mg - Mg = \dfrac{M {v}^{2} }{r}}$

$\sf{ \implies \: Mg = \dfrac{M {v}^{2} }{r}} \: ......(2)$

Finally putting value of (1) and (2) in the 1st equation , we get :

$\boxed{ \sf{Mgh = \frac{1}{2}M {v}^{2} \bigg\{1 + \frac{ {k}^{2} }{ {R}^{2} } \bigg\}}}$

$\sf{ \implies \:( M \dfrac{ {v}^{2} }{r}) h = \dfrac{1}{2}M {v}^{2} \bigg\{1 + \beta \bigg\}}$

$\sf{\implies \: 1 + \beta = \dfrac{2h}{r} }$

$\sf{\implies \: \beta = \bigg( \dfrac{2h}{r} - 1 \bigg)}$

So final answer is :

$\boxed{ \red{ \bold{ \huge{ \sf{\: \beta = \bigg( \dfrac{2h}{r} - 1 \bigg)}}}}}$