Question

A spherical ball is kept at the edge of a table a shown. In case-I, the table & edge is smooth. In case – II, the edge is rough so that ball does not slip on the edge. In both cases ball is unstable and leaves the edge after rotating by a certain angle ($\theta$) about edge. It’s velocity of mass at the time of leaving is v. Then,

• A. ${\theta }_1>{\theta }_2$
• B. ${\theta }_1<{\theta }_2$
• C. $v_1>v_2$
• D. $v_1=v_2$

Answer 1

Answer: (B), (C)

The correct options are
B ${\theta }_1<{\theta }_2$
C $v_1>v_2$

Applying law of conservation of energy in case of rolling:
$mg(R-Rcos\frac{\theta }{2})=\frac{1}{2}m{v}_2^2+\frac{1}{2}\times \frac{2}{5}m{R}^2\times \frac{v_0^2}{R^2}$
$v_2=\sqrt{\frac{10}{7}gR(1-cos{\theta }_2)}$
Also. mg $cos{\theta }_2$ $=\frac{m{v}_2^2}{R}=\frac{10g}{7}(1-cos{\theta }_2)$. Hence,
17 cos ${\theta }_2$ = 10
$cos{\theta }_2=\frac{10}{17}$
In case of frictionless surface:
$mg(R-Rcos{\theta }_1)=\frac{1}{2}m{v}_1^2$
$v_1=\sqrt{2gR(1-cos{\theta }_1)}$ and
$mgcos{\theta }_1=\frac{m{v}_1^2}{R}=2mg(1-cos{\theta }_1)cos{\theta }_1=\frac{2}{3}.Hencecos{\theta }_2<cos{\theta }_1\Rightarrow cos{\theta }_2>{\theta }_1{v}_1=\sqrt{2gR(1-\frac{2}{3})}=\sqrt{\frac{2gR}{3}}{v}_2=\sqrt{\frac{10gR}{7}\times \frac{7}{17}}$