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System of Particles and Rotational Motion

Torque

A particle of mass m is moving along a trajectory given by x=x

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Solving time: 3 mins

A particle of mass $m$ is moving along a trajectory given by
$x = x_{0} + a cos ω_{1} t$ and $y = y_{0} + b sin ω_{2} t$ . The torque acting on the particle about the origin at $t = 0$ is
(a) zero
(b) $m (- x_{0} b + y_{0} a) ω_{1}^{2} \hat{k}$
(c) $- m (x_{0} b ω_{2}^{2} - y_{0} a ω_{1}^{2}) \hat{k}$
(d) $+ m y_{0} a ω_{1}^{2} \hat{k}$

Views: 5,968 students

Updated on: Sep 27, 2022

Book:

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

Exam:

JEE Mains 2019

Found 5 tutors discussing this question

Asher Discussed

A particle of mass

m

is moving along a trajectory given by

x = x_{0} + a cos ω_{1} t

and

y = y_{0} + b sin ω_{2} t

. The torque acting on the particle about the origin at

t = 0

is
(a) zero
(b)

m (- x_{0} b + y_{0} a) ω_{1}^{2} \hat{k}

(c)

- m (x_{0} b ω_{2}^{2} - y_{0} a ω_{1}^{2}) \hat{k}

(d)

+ m y_{0} a ω_{1}^{2} \hat{k}

8 mins ago

Discuss this question LIVE

8 mins ago

Text solutionVerified

Exp. (d)
Given equations of trajectory are

x = x_{0} + a cos ω_{f}

y = y_{0} + b sin ω_{2} t

Now,

v_{x} = d x / d t = - a ω_{1} sin ω_{f}

and

v_{y} = \frac{d y}{d t} = b ω_{2} cos ω_{2} t

and
Similarly,

\Rightarrow a_{x} = \frac{d v _{x}}{d t} = - a ω_{1}^{2} cos (ω, t)

Similarly,

a_{y} = \frac{d v _{y}}{d t} = - b ω_{2}^{2} sin (ω_{2} t)

Now, at

t = 0; x = x_{0} + a; y = y_{0}

and

a_{x} = - a ω_{1}^{2}; a_{y} = 0

Torque acting on the particle is given as

τ = r \times F = m (r \times a)

τ = m (- a ω_{1}^{2}) \times y_{0} (- \hat{k})

Alternate Solution
We can calculate total force from Eq. (i),

F = - m (a ω_{1}^{2} cos ω_{f} \hat{i} + b ω_{2}^{2} sin ω_{2} + \hat{j})

and

r = x \hat{i} + \hat{y j} = (x_{0} + acos ω \overset{}{^}) \hat{i}

+ (y_{0} + b sin ω_{2} t) \hat{j}

Now, torque,

τ = r \times F

= [(x_{0} + a cos ω f) \hat{i} + (y_{0} + b sin ω_{2} t) \hat{j}]

\times [- m (a ω_{1}^{2} cos ω_{1} t \hat{i} + b ω_{2}^{2} sin ω_{2} + \hat{j})]

= - m ⎣ ⎡ \hat{i} x_{0} + a cos ω_{t} a ω_{1}^{2} cos ω_{t} \hat{j} y_{0} + b sin ω_{2} t b ω_{2}^{2} sin ω_{2} t \hat{k} 00 ⎦ ⎤

= - m [x_{0} b ω_{2}^{2} sin ω_{2} t + ab ω^{2} sin ω_{2} t cos ω_{1} t

- y_{0}

ω_{1}^{2} cos ω t

- ab ω_{1}^{2} sin ω_{2} + cos ω_{1} t] \hat{k}

On solving it at

t = 0

, we get

τ = ma ω_{1}^{2} y_{0} \hat{k}

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Question Text	A particle of mass $m$ is moving along a trajectory given by $x = x_{0} + a cos ω_{1} t$ and $y = y_{0} + b sin ω_{2} t$ . The torque acting on the particle about the origin at $t = 0$ is (a) zero (b) $m (- x_{0} b + y_{0} a) ω_{1}^{2} \hat{k}$ (c) $- m (x_{0} b ω_{2}^{2} - y_{0} a ω_{1}^{2}) \hat{k}$ (d) $+ m y_{0} a ω_{1}^{2} \hat{k}$
Updated On	Sep 27, 2022
Topic	System of Particles and Rotational Motion
Subject	Physics
Class	Class 11
Answer Type	Text solution:1 Video solution: 6
Upvotes	731
Avg. Video Duration	10 min

Text solutionVerified

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