Question

just before the sphere 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 C $\frac{3mg}{2}$

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

Before coming in contact with the peg, the sphere rotates in a circle of radius $r=0.8$ m

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

From figure, $PC=rsin{30}^o=\frac{r}{2}$ 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 $mg\times \frac{r}{2}=\frac{m{v}^2}{2}-0$

$⟹\frac{m{v}^2}{r}=mg$

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

$\therefore$ $mg=T-\frac{mg}{2}$

$⟹T=\frac{3mg}{2}$