JEE Main 201910 AprilMorning Shift
- Exam
- JEE Main
- Year
- 2019
- Date
- 10 April
- Shift
- Morning Shift
- Subject
- Physics
- Chapter
- Motion in Two Dimensions
JEE Main 2019 · 10 April · Morning Shift · Physics · Motion in Two Dimensions
A particle of mass m is moving along a trajectory given by x = x _...
A particle of mass is moving along a trajectory given by
The torque, acting on the particle about the origin, at is:
The torque, acting on the particle about the origin, at is:
Options
- A+ m{y}_{0}a{\omega}_{1}^{2}\overset{^}{k}
- B- m\left({x}_{0}b{\omega}_{2}^{2} - {y}_{0}a{\omega}_{1}^{2}\right)\overset{^}{k}
- CZero
- Dm\left(- {x}_{0}b + {y}_{0}a\right){\omega}_{1}^{2}\overset{^}{k}
Correct answer
A. + m{y}_{0}a{\omega}_{1}^{2}\overset{^}{k}
Step-by-step solution
At ,
\overset{\to}{F} = m\overset{\to}{a} = \left(- a{\omega}_{1}^{2}\overset{^}{i}\right)
And
\overset{\to}{r} = \left({x}_{0}\text{+}a\right)\overset{^}{i} + {y}_{0}\overset{^}{j}
Torque,
τ = \left(\left({x}_{0}\text{+}a\right)\overset{^}{i} + {y}_{0}\overset{^}{j}\right) \times \left(- a{\omega}_{1}^{2}\overset{^}{i}\right)
= m{y}_{0}a{\omega}_{1}^{2}\overset{^}{k}