SATHEE: Rotational Motion Question 141

Rotational Motion Question 141

Question: A pulley of radius 2 m is rotated about its axis by a force $F = (20 t - 5 t^{2})$ newton (where $t$ is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is $10 k g m^{2}$ , the number of rotations made by the pulley before its direction of motion is reversed, is:

Options:

A) less than 3

B) more than 3 but less than 6

C) more than 6 but less than 9

D) more than 9

Show Answer

Answer:

Correct Answer: B

Solution:

[b]

$τ = (20 t - 5 t^{2}) 2 = 40 t - 10 t^{2}$

$α = \frac{τ}{I} = \frac{40 t - 10 t^{2}}{10} = 4 t - t^{2}$

$ω = \int_{0}^{t} α d t = 2 t^{2}, - \frac{t^{3}}{3}$

When direction is reversed, co is zero.

So,

$2 t^{2} - \frac{t^{3}}{3} = 0 \Rightarrow t^{3} = 6 t^{2} \Rightarrow t = 6 s$

$θ = \int ω d t$

$= \int_{0}^{6} (2 t^{2} - \frac{t^{3}}{3}) d t$

$= {\frac{2 t^{2}}{3} - \frac{t^{4}}{12}}^{6}_{0} = 36 r a d$

Number of revolution = $\frac{36}{2 π}$ = Less than 6

Rotational Motion Question 141

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Rotational Motion Question 141

Options:

Answer:

Solution:

Engineering Preparation

Medical Preparation

NCERT Books & Solutions

Important Resources

Other Resources

Syllabus

Test Platform

Sample Mock Test

Mentorship

NCERT Exercise Video Solution

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