Question

91. A particle of mass m is released on a smooth
track in vertical plane which transforms into a
circular arc of radius R. Find the reaction force
exerted at the bottom most position.
m
3R
R
(1) 6mg
(2) 7mg
(3) 5mg
(4) 9mg​

Answer 1

Answer:

6mg

Explanation:

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Answer 2

Answer:

The correct answer is equal to option(2).

Explanation:

The Kinetic energy of a particle at the highest point is denoted by $KE_{h}$.

The potential energy of a  particle  at the highest point is denoted by $PE_{h}$.

The Kinetic energy of a particle at the bottom-most position is denoted by $KE_{b}$.

The potential energy of a particle at the bottom-most position is denoted by $PE_{b}$.

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

$KE_{h} +PE_{h}= KE_{b} +PE_{b}$

$0+mg(3R)=\frac{1}{2}mv^{2}+0$

$v=\sqrt{6gR}$

The reaction force at the bottom-most position,

N = weight of particle + centripetal force

   = mg + $\frac{mv^{2} }{R}$

   = mg + $\frac{m(\sqrt{6gR}) ^{2} }{R}$

   = 7mg

So, the reaction force exerted at the bottom most position is equal to 7mg.