Question

A body of mass $m=2\text{kg}$, initially at rest and sliding along a frictionless track from a height $\text{h}$ (as shown in the figure) just completes a vertical circle of diameter $\text{AB= 3 m}$. The height $\text{h}$ is equal to

• A. $2.4\text{m}$
• B. $4\text{m}$
• C. $3.75\text{m}$
• D. $5\text{m}$

Answer 1

The correct option is C $3.75\text{m}$


Given
$\text{m = 2 kg}$
$\text{AB, D = 3 m}$
$\text{h}$= ?

Applying law of conservation of energy at point A & P:
$P{E}_A+K{E}_A=P{E}_P+K{E}_P$
$\frac{1}{2}m{v}^2=mgh$
$v_A=\sqrt{2gh}.....\left(i\right)$

Datum is at A $\therefore P{E}_A=0$
Also, $v_P=0,K{E}_p=0$
To just complete vertical circle, speed at point A
$=\sqrt{5g\times \left(D/2\right)}.....\left(ii\right)$
Equating (i) & (ii),
$\sqrt{2gh}=\sqrt{5g\times \left(D/2\right)}$
$h=\frac{5D}{4}=\frac{5}{4}\times 3$
$h=3.75\text{m}$

Hence option C is the correct answer