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System of Particles and Rotational Motion

Moment of inertia

A metal coin of mass 5 {~g} and radius 1 {~cm} is fixed to a t

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Hard

Solving time: 4 mins

A metal coin of mass $5 g$ and radius $1 cm$ is fixed to a thin stick $A B$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about $A B$ at 25 rotations per second in 5 s, is close to

(a) $4.0 \times 1 0^{- 6} N - m$
(b) $2.0 \times 1 0^{- 5} N - m$
(c) $1.6 \times 1 0^{- 5} N - m$
(d) $7.9 \times 1 0^{- 6} N - m$

Views: 5,356 students

Book:

Arihant Physics JEE Main Chapterwise Solutions (2019-2002) (Arihant)

Exam:

JEE Mains 2019

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Daniel Discussed

A metal coin of mass

5 g

and radius

1 cm

is fixed to a thin stick

A B

of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about

A B

at 25 rotations per second in 5 s, is close to

(a)

4.0 \times 1 0^{- 6} N - m

(b)

2.0 \times 1 0^{- 5} N - m

(c)

1.6 \times 1 0^{- 5} N - m

(d)

7.9 \times 1 0^{- 6} N - m

11 mins ago

Discuss this question LIVE

11 mins ago

Text solutionVerified

Exp. (b)
Moment of inertia (MI) of a disc about a tangential axis in the plane of disc can be obtained as below.
Solution Image

Moment of inertia of disc about it's axis,

l_{1} = \frac{M R ^{2}}{2}

From perpendicular axes theorem, moment of inertia of disc about an axis along it's diameter is

I_{x} + I_{y} = I_{z} \Rightarrow 2 I_{2} = I_{1}

I_{2} = \frac{I _{1}}{2} = \frac{M R ^{2}}{4}

I_{y} = I_{2}

So, moment of inertia about a about a tangential axis from parllel axes theorem is

= \frac{5}{4} M R^{2}

Now, using torque,

τ = l α,

we have

τ = I α = \frac{5}{4} M R^{2} (\frac{ω _{f} - ω _{j}}{Δ t})

Here,

M = 5 \times 1 0^{- 3} kg, R = 1 \times 1 0^{- 2} m

ω_{f} = 25 rps = 25 \times 2 π \frac{rad}{s} = 50 π \frac{rad}{s}

ω_{i} = 0, \frac{Δ t}{5} = 5 s

So, τ = \frac{\frac{5}{4} \times 5 \times 1 0 ^{- 3} \times ( 1 0 ^{- 2} ) ^{2} \times 50 π}{5}

\propto 2 \times 1 0^{- 5} N - m

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Question Text	A metal coin of mass $5 g$ and radius $1 cm$ is fixed to a thin stick $A B$ of negligible mass as shown in the figure. The system is initially at rest. The constant torque, that will make the system rotate about $A B$ at 25 rotations per second in 5 s, is close to (a) $4.0 \times 1 0^{- 6} N - m$ (b) $2.0 \times 1 0^{- 5} N - m$ (c) $1.6 \times 1 0^{- 5} N - m$ (d) $7.9 \times 1 0^{- 6} N - m$
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1
Upvotes	25

Text solutionVerified

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