Question

A block is freely sliding down from a vertical height $4m$ on a smooth inclined plane. The block reaches bottom of inclined plane then it describes vertical circle of radius $1m$ along smooth track. The ratio of normal reactions on the block while it is crossing lowest point and highest point of vertical circle is:

• A. $6:1$
• B. $5:1$
• C. $3:1$
• D. $5:2$

Answer 1

The correct option is C $3:1$

applying the law of conservation of energy at point A and B, we have
$mgh=\frac{1}{2}m{v_1}^2\Rightarrow v_1=\sqrt{2gh}\Rightarrow v_1=\sqrt{8g}$
Now, net force is towards the center is the centripetal force, we have at point B
$\frac{m{v_1}^2}{r}=N_1-mg$
$\Rightarrow N_1=\frac{m{v_1}^2}{r}+mg=\frac{m\left(8g\right)}{1}+mg=9mg$ ..........$\left(I\right)$
applying the law of conservation of energy at point B and C, we have
$\frac{1}{2}m{v_1}^2=mg\left(2r\right)+\frac{1}{2}m{v_2}^2\Rightarrow \frac{1}{2}m\left(8g\right)=mg\left(2\right)+\frac{1}{2}m{v_2}^2\Rightarrow v_2=\sqrt{4g}$
Now, net force is towards the center is the centripetal force, we have at point C
$\frac{m{v_2}^2}{r}=N_2+mg$
$\Rightarrow N_2=\frac{m{v_2}^2}{r}-mg=\frac{m\left(4g\right)}{1}-mg=3mg$ ..........$\left(II\right)$
So the ratio of normal reactions on the block while it is crossing lowest point, highest point of vertical circle is $\frac{N_1}{N_2}=\frac{9mg}{3mg}=3:1$ ..from $\left(I\right)$ and $\left(II\right)$