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Law of Conservation of Mechanical Energy

A particle is...

Question

A particle is released from rest from a height of $4 h$ , on a smooth track as shown in figure. When the particle reaches point $C$ , then the normal reaction on it by the track is:
( $C$ is the highest point on the circular part of the track)

m g

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3 m g

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4 m g

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\frac{2}{3} m g

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Solution

The correct option is B $3 m g$
Applying mechanical energy conservation from $A$ to $C$ ,
$Loss in P.E = Gain in K.E = K E_{f} - K E_{i}$
$m g (2 h) = \frac{1}{2} m v^{2} - 0$
$∴ v^{2} = 4 g h . . . (i)$

Applying equation of dynamics at point $C$ , from the FBD of particle:
$N + m g = \frac{m v^{2}}{r} . . . (i i)$
where $r = h$

From Eq $(i) & (i i)$ :
$N + m g = \frac{m (4 g h)}{h}$
$∴ N = 4 m g - m g = 3 m g$

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Law of Conservation of Mechanical Energy

Standard XII Physics

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A particle is released from rest from a height of 4h, on a smooth track as shown in figure. When the particle reaches point C, then the normal reaction on it by the track is: (C is the highest point on the circular part of the track)

The correct option is B 3mgApplying mechanical energy conservation from A to C, Loss in P.E=Gain in K.E=KEf−KEi mg(2h)=12mv2−0 ∴v2=4gh ...(i) Applying equation of dynamics at point C, from the FBD of particle: N+mg=mv2r ...(ii) where r=h From Eq (i) & (ii): N+mg=m(4gh)h ∴N=4mg−mg=3mg

A particle is released from rest from a height of $4 h$ , on a smooth track as shown in figure. When the particle reaches point $C$ , then the normal reaction on it by the track is:
( $C$ is the highest point on the circular part of the track)