Physics

System of Particles and Rotational Motion

Rotational dynamics

A small sphere of mass { m }, is attached to a rod of negligi

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Solving time: 3 mins

A small sphere of mass $m$ , is attached to a rod of negligible mass and length $1 m$ (upto centre of sphere) and kept on a rough floor in a vertical position. The sphere is given a slight push to right side so that it starts falling (a) The frictional forces developed on the rod initially will be directed to left. (b) The frictional forces developed on the rod initially will be directed to right (c) The normal reaction $N$ will be zero when the rod has fallen through an angle $θ = cos^{- 1} \frac{2}{3}$ with the vertical. (d) The normal reaction $N$ will be zero when the rod has fallen through an angle $θ = sin^{- 1} \frac{3}{5}$ with the vertical.

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Book:

IIT JEE Super Course in Physics Mechanics II (Pearson)

Text solutionVerified

If there is no friction, the COM will be falling vertically down, i.e., bottom end will be moving to left. Hence, when there is friction, the direction of friction is such as to oppose the tendency of the bottom end to move to the left and so it will act as shown, i.e., towards right. Resolving forces along radial

\Rightarrow mg cos θ - m ω^{2} ℓ

= N cos θ + f sin θ - (1)

From energy equation

ℓ (1 - cos θ) mg

= \frac{1}{2} I ω^{2} = \frac{1}{2} m ℓ^{2} ω^{2}

∴ ω^{2} = \frac{2 g}{ℓ} (1 - cos θ)

\Rightarrow

substitute (2) in (1)

mg cos θ - m \frac{2 g}{ℓ} (1 - cos θ) \cdot ℓ

= N cos θ + f sin θ

f = 0

, when

N = 0

For

N

to be zero

\Rightarrow

LHS

= 0 \Rightarrow mg cos θ = 2 mg

(1 - cos θ)

cos θ = \frac{2}{3}

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Harper Discussed

A small sphere of mass

m

, is attached to a rod of negligible mass and length

1 m

(upto centre of sphere) and kept on a rough floor in a vertical position. The sphere is given a slight push to right side so that it starts falling (a) The frictional forces developed on the rod initially will be directed to left. (b) The frictional forces developed on the rod initially will be directed to right (c) The normal reaction

N

will be zero when the rod has fallen through an angle

θ = cos^{- 1} \frac{2}{3}

with the vertical. (d) The normal reaction

N

will be zero when the rod has fallen through an angle

θ = sin^{- 1} \frac{3}{5}

with the vertical.

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Question Text	A small sphere of mass $m$ , is attached to a rod of negligible mass and length $1 m$ (upto centre of sphere) and kept on a rough floor in a vertical position. The sphere is given a slight push to right side so that it starts falling (a) The frictional forces developed on the rod initially will be directed to left. (b) The frictional forces developed on the rod initially will be directed to right (c) The normal reaction $N$ will be zero when the rod has fallen through an angle $θ = cos^{- 1} \frac{2}{3}$ with the vertical. (d) The normal reaction $N$ will be zero when the rod has fallen through an angle $θ = sin^{- 1} \frac{3}{5}$ with the vertical.
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	138

Text solutionVerified

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