Question

just after it comes in contact with the peg.

• A. $\frac{mg}{2}$
• B. $mg$
• C. $\frac{3mg}{2}$
• D. $\frac{5mg}{2}$

Answer 1

The correct option is D $\frac{5mg}{2}$

Given : $r=0.8$ m $u=0$ m/s $OA=0.4$ m

$\therefore$ $AM=r^{\prime }=0.8-0.4=0.4$ m

After coming in contact with the peg, the sphere rotates in a circle of radius $r^{\prime }=0.4$ m

Let the velocity of the sphere when it reaches the point v be $v$.

From figure, $PC=rsin{30}^o=0.4$ m

Work-energy theorem : $W=\Delta K.E$

$\therefore$ $mg\left(PC\right)=\frac{1}{2}m{v}^2-\frac{1}{2}m{u}^2$

OR $mg\times 0.4=\frac{m{v}^2}{2}-0$

$⟹m{v}^2=0.8mg$

Using: $m{a}_r=\frac{m{v}^2}{r^{\prime }}=T^{\prime }-mgsin{30}^o$

$\therefore$ $\frac{0.8mg}{0.4}=T^{\prime }-\frac{mg}{2}$

$⟹T^{\prime }=\frac{5mg}{2}$