Question

A particle of mass $m$ is released on a smooth track in a vertical plane which transforms into a circular arc of radius $R$. Find the reaction force exerted at the bottom-most position.

• A. $6mg$
• B. $7mg$
• C. $5mg$
• D. $9mg$

Answer 1

The correct option is A

$6mg$

Step 1. Given Data,

Height,$h$= $3R$

$R$is the radius of the circular arc

$m$is the mass of the particle

Step 2. Formula used,

Potential energy, $PE=mgh$

Kinetic energy, $KE=\frac{1}{2}m{v}^2$

Centripetal force, $N=\frac{m{v}^2}{R}$

$h$is the height

$R$is the radius of the circular arc

$v$ is the velocity of the particle

$m$is the mass of the particle

Principle of conservation of energy,

$K{E}_{initial}+U_{initial}=K{E}_{final}+U_{final}$, which we can write as $K{E}_h+P{E}_h=K{E}_b+P{E}_b$

The Kinetic energy of a particle at the highest point is $K{E}_h$.

The potential energy of a particle at the highest point is $P{E}_h$.

The Kinetic energy of a particle at the bottom-most position is $K{E}_b$.

The potential energy of a particle at the bottom-most position is $P{E}_h$.

Step 3. Calculations of the force,

Apply the principle of conservation of energy from the highest point to the bottom-most position,

Height, $h$= $3R$

$$\begin{gathered} K{E}_h+P{E}_h=K{E}_b+P{E}_b \\ 0+mg\left(3R\right)=\frac{1}{2}m{v}^2+0 \\ v=\sqrt{6gR} \end{gathered}$$

The reaction force at the bottom-most position,

N = centripetal force

$$\begin{gathered} =\frac{m{v}^2}{R} \\ =\frac{m({\sqrt{6gR)}}^2}{R} \\ =6mg \end{gathered}$$

Therefore the reaction force at the bottom-most position is $6mg$.

The reaction force is the effect of inertia, not due to actual forces. As as for the centripetal force, its reaction force is just from the object acting on whatever is supplying the centripetal force.

Hence option A is the correct answer.