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Question: A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is \(2 \m...

A particle is moving on a circular path of 10 m radius. At any instant of time, its speed is 2 m s−22 \mathrm {~m} \mathrm {~s} ^ { - 2 } . The force acting on the particle is :

A

mω2r→m \omega ^ { 2 } \overrightarrow { \mathbf { r } }

B

−mω2r→- m \omega ^ { 2 } \overrightarrow { \mathrm { r } }

C

2mω2r→2 m \omega ^ { 2 } \overrightarrow { \mathbf { r } }

D

−2mω2r→- 2 m \omega ^ { 2 } \overrightarrow { \mathrm { r } }

Answer

−mω2r→- m \omega ^ { 2 } \overrightarrow { \mathrm { r } }

Explanation

Solution

Given r⃗=Acos⁡ωti^+Bsin⁡ωtj^\vec { r } = A \cos \omega t \hat { i } + B \sin \omega t \hat { j }

Velocity, v⃗=dr⃗dt=ddt(Acos⁡ωti^+Bsin⁡ωtj^)\vec { v } = \frac { d \vec { r } } { d t } = \frac { d } { d t } ( A \cos \omega t \hat { i } + B \sin \omega t \hat { j } )

=−Aωsin⁡ωti^+Bωcos⁡ωtj^= - A \omega \sin \omega t \hat { i } + B \omega \cos \omega t \hat { j }

Acceleration,

a⃗=dv⃗dt=−Aω2cos⁡ωti^−Bω2sin⁡ωtj^\vec { a } = \frac { d \vec { v } } { d t } = - A \omega ^ { 2 } \cos \omega t \hat { i } - B \omega ^ { 2 } \sin \omega t \hat { j }

a⃗=ω2[Acos⁡ωti^+Bsin⁡ωtj^]=−ω2r⃗\vec { a } = \omega ^ { 2 } [ A \cos \omega t \hat { i } + B \sin \omega t \hat { j } ] = - \omega ^ { 2 } \vec { r }

The force acting on the particle is

F⃗=ma⃗=m(−ω2r⃗)=−mω2r⃗\vec { F } = m \vec { a } = m \left( - \omega ^ { 2 } \vec { r } \right) = - m \omega ^ { 2 } \vec { r }