Question

Figure (10 - E15) 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 radial and the tangential accelerations of the centre when the ball is at A.

Answer 1

(a) Total kinetic energy,

$y=\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

Therefore, according to the question

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

$\Rightarrow mgH-mgR(1+sin\theta )$

=$\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

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

(b) To find the acceleration components,

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

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

$\Rightarrow v^2=\frac{10}{7}g(H-R-Rsin\theta )...\left(i\right)$

$\Rightarrow \frac{v^2}{R}=\frac{10}{7}\frac{g\left[\right(H-R)-Rsin\theta ]}{R}$

$\to$ radial acceleration

$\Rightarrow \frac{v^2}{R}=\frac{10}{7}\frac{g\left[\right(H-R)-Rsin\theta ]}{R}$

$\Rightarrow 2v\frac{dv}{dt}=-\left(\frac{10}{7}\right)gRcos\theta \frac{d\theta }{dt}$

$\Rightarrow \omega R\frac{dv}{dt}=-\left(\frac{5}{7}\right)gRcos\theta \frac{d\theta }{dt}$

$\Rightarrow \frac{dv}{dt}=-\left(\frac{5}{7}\right)gcos\theta$

$\to$ tangential acceleration

(c) Normal force at $\theta$ = 0, $\frac{m{v}^2}{R}=\left(\frac{70}{1000}\right)\times \left(\frac{10}{7}\right)\times 10\left\{\frac{(0.6-0.1)}{0.1}\right\}=5N$

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

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

$=0.07(10-\frac{5}{7}\times 10)$

$=\frac{1}{100}(70-50)=0.2N$