Question

Figure shows a small spherical ball of mass m rolling down the loop track. The ball is released on the linear portion at a vertical height $H$ from the lowest point. The circular part shown has a radius $R$.
(a) Find the kinetic energy of the ball when it is at a point A where the radius makes an angle $\theta$ with the horizontal.
(b) Find the radius and the tangential accelerations of the centre when the ball is at $A$.
(c) Find the normal force and the frictional force acting on the ball if $H=60cm,R=10cm,\theta =0$ and $m=70g$.

Answer 1

a) The change in the potential energy is the sum of linear kinetic energy and the rotational kinetic energy of the body:

$mgh=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+mg{h}^{\prime }$

Therefore according to the question

$mgH=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2+mgR(1+\sin \theta )$

$\Rightarrow mgH-mgR(1+\sin \theta )=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Therefore, the total kinetic energy is the sum of linear kinetic energy and the rotational kinetic energy:

$\Rightarrow KE=mg(H-R-R\sin \theta )$

b) To find the acceleration components

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=mg(H-R-R\sin \theta )$

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\times \frac{2}{5}\times m{R}^2\times \frac{v^2}{R^2}=mg(H-R-R\sin \theta )$

$\Rightarrow \frac{7}{10}m{v}^2=mg(H-R-R\sin \theta )$

Simplify the above expression:

$\frac{v^2}{R}=\frac{10g}{7R}[H-R-R\sin \theta ]$

$\Rightarrow$ radial acceleration$\Rightarrow 2v\frac{dv}{dt}=-\frac{10}{7}gR\cos \theta \frac{d\theta }{dt}$

$\Rightarrow \omega R\frac{dv}{dt}=-\frac{5}{7}gR\cos \theta \frac{d\theta }{dt}$

$\Rightarrow \frac{dv}{dt}=-\frac{5}{7}g\cos \theta \to$ tangential acceleration.

c) Normal force at the point A is $=0$

$\Rightarrow F=m{a}_r=\frac{70}{1000}\times \frac{10}{7}\times 10\left(\frac{0.6-0.1}{0.1}\right)=5N$

Frictional force is given as:

$f=mg-ma=m(g-a)$

$=m(10-\frac{5}{7}\times 10)$

$=0.07\left(\frac{70-50}{7}\right)=\frac{1}{100}\times 20=0.2N$