A smooth inclined plane ends in a vertical circular loop, as shown in the figure

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 $h = α R$ . The value of $α$ is [2026]

Smart Achievers · Accepted Answer

(2)

$At highest point$

$m g + N = \frac{m v_{top}^{2}}{R}$

$m g + 3 m g = \frac{m v_{top}^{2}}{R} \Rightarrow v_{top} = \sqrt{4 g R}$

$Now C.O.M.E:$

$m g (h - 2 R) = \frac{1}{2} m \cdot 4 g R \Rightarrow h = 4 R$

Solution

Q.
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 $h = α R$ . The value of $α$ is [2026]

1 2

2 4

3 3

4 6

Ans.
(2)

$At highest point$

$m g + N = \frac{m v_{top}^{2}}{R}$

$m g + 3 m g = \frac{m v_{top}^{2}}{R} \Rightarrow v_{top} = \sqrt{4 g R}$

$Now C.O.M.E:$

$m g (h - 2 R) = \frac{1}{2} m \cdot 4 g R \Rightarrow h = 4 R$

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Solution

Q. 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=αR. The value of α is [2026]

1 2

2 4

3 3

4 6

Ans. (2) At highest point mg+N=mvtop2R mg+3mg=mvtop2R⇒vtop=4gR Now C.O.M.E: mg(h-2R)=12m·4gR⇒h=4R

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Q.
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 $h = α R$ . The value of $α$ is [2026]

Ans.
(2)

$At highest point$

$m g + N = \frac{m v_{top}^{2}}{R}$

$m g + 3 m g = \frac{m v_{top}^{2}}{R} \Rightarrow v_{top} = \sqrt{4 g R}$

$Now C.O.M.E:$

$m g (h - 2 R) = \frac{1}{2} m \cdot 4 g R \Rightarrow h = 4 R$