Question 26: Work, Energy and Power & Circular Motion

Question

A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height $h$ as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height $h=\alpha R$. The value of $\alpha$ is $\_\_\_\_$

Accepted Answer

The potential energy at height $h$ converts into kinetic energy as the object slides down.
To move in a circle of radius $R$, the net force toward the center must be $\frac{m v^2}{R}$.
Let the bottom of the loop be the reference level (Potential Energy $=0$ ).
At release point, the gravitational potential energy is $\mathrm{PE}_1=\mathrm{mgh}$, and the kinetic energy is $\mathrm{KE}_1=0$.
So, total energy of the body at releasing point is $\mathrm{E}_1=\mathrm{PE}_1+\mathrm{KE}_1=\mathrm{mgh}+0=\mathrm{mgh}$
At the highest point of circle the height is 2 R (the diameter).
So, the gravitational potential energy is $\mathrm{PE}_2=\mathrm{mg}(2 \mathrm{R})$.
Let velocity at the top be $\mathrm{v}_2$. So, the kinetic energy of body at top is $\mathrm{KE}_2=\frac{1}{2} \mathrm{mv}_2^2$.
So, the total energy of the body at top of the circle is $\mathrm{E}_2=2 \mathrm{mgR}+\frac{1}{2} \mathrm{mv}_2^2$.
Using law of conservation of energy :
$$ \mathrm{mgh}=2 \mathrm{mgR}+\frac{1}{2} \mathrm{mv}_2^2 $$
$$ g h=2 g R+\frac{v_2^2}{2} \Rightarrow v_2^2=2 g(h-2 R) \ldots (i)$$
At the top of the loop, normal force ( N ) acts on the body from the track.
It is given that the body exerts a force of 3 mg on the plane, so by Newton's 3rd Law, the track exerts $\mathrm{N}=3 \mathrm{mg}$ on the body.
So, total centripetal force is $\mathrm{N}+\mathrm{mg}=\frac{\mathrm{mv}_2^2}{\mathrm{R}}$
Substituting N = 3mg:
$$ 3 \mathrm{mg}+\mathrm{mg}=\frac{\mathrm{mv}_2^2}{\mathrm{R}} \Rightarrow 4 \mathrm{mg}=\frac{\mathrm{mv}_2^2}{\mathrm{R}} $$
$$ \mathrm{v}_2^2=4 \mathrm{gR} \ldots 	ext { (ii) } $$
From both the equations,
$$ 2 \mathrm{~g}(\mathrm{~h}-2 \mathrm{R})=4 \mathrm{gR} $$
$$\Rightarrow $$ $h-2 R=2 R$
$$\Rightarrow $$ $$ h=4 R=\alpha R \Rightarrow \alpha=4 $$
Therefore, the value of $\alpha$ is 4 . Thus, the correct option is (B).

Answer

2

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4

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3

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6

Work, Energy and Power & Circular Motion

A smooth inclined plane ends in a vertical circular loop, as shown in the figure. A small body is released from height h as shown. If the body exerts a force of three times its weight on the plane at the highest point of circle then the height h=\alpha R. The value of \alpha is \_\_\_\_

Choose the correct answer: