Solveeit Logo
Login

Question

Question: Two particles describe the same particles of circle \(R\) in the same direction with the same speed ...

Two particles describe the same particles of circle RR in the same direction with the same speed VV, then at given instant the relative angular velocity of 22 with respect to 11 will be given as,

A.2Vsin⁡θ2R B.V2Rsin⁡θ2 C.VR D.Vcos⁡θ2R \begin{aligned} & A.\dfrac{2V\sin \dfrac{\theta }{2}}{R} \\\ & B.\dfrac{V}{2R\sin \dfrac{\theta }{2}} \\\ & C.\dfrac{V}{R} \\\ & D.\dfrac{V\cos \dfrac{\theta }{2}}{R} \\\ \end{aligned}

Explanation

Solution

First of all find out the velocity at each point and take the relative velocity of 22 with respect to 11. The angular velocity can be found by taking the ratio of the relative velocity of the particle to the radius of the circle. Using this find the angular velocity and arrive at its magnitude. Simplify the equation and find out the correct answer. This will help you in answering this question.

Complete step by step solution:
The velocity at 22 is given as,
V2=(vcos⁡θ)j^−vsin⁡θi^{{V}_{2}}=\left( v\cos \theta \right)\hat{j}-v\sin \theta \hat{i}
The velocity at 11 has been mentioned as,
V1=vj^{{V}_{1}}=v\hat{j}
The relative velocity can be found by taking the difference of the velocity at 22 and the velocity at 11. This can be shown as,
V21=V2−V1{{V}_{21}}={{V}_{2}}-{{V}_{1}}
Substituting the values in it will give,
V21=(vcos⁡θ)j^−vsin⁡θi^−vj^{{V}_{21}}=\left( v\cos \theta \right)\hat{j}-v\sin \theta \hat{i}-v\hat{j}
The angular velocity can be found by taking the ratio of the relative velocity to the radius of the circle,
ω=V21R=(vcos⁡θ−v)j^−vsin⁡θi^R\omega =\dfrac{{{V}_{21}}}{R}=\dfrac{\left( v\cos \theta -v \right)\hat{j}-v\sin \theta \hat{i}}{R}
The magnitude of the angular velocity can be found as,
ω=v(cos⁡θ−1)2+(−sin⁡θ)2R\omega =\dfrac{v\sqrt{{{\left( \cos \theta -1 \right)}^{2}}+{{\left( -\sin \theta \right)}^{2}}}}{R}
Simplifying this value can be shown as,
ω=vcos⁡2θ+1−2cos⁡θ+sin⁡2θR\omega =\dfrac{v\sqrt{{{\cos }^{2}}\theta +1-2\cos \theta +{{\sin }^{2}}\theta }}{R}
That is we can write that,
ω=vR2−2cos⁡θ ⇒ω=vR2(1−cos⁡θ) ⇒ω=vR2×2sin⁡2θ2 \begin{aligned} & \omega =\dfrac{v}{R}\sqrt{2-2\cos \theta } \\\ & \Rightarrow \omega =\dfrac{v}{R}\sqrt{2\left( 1-\cos \theta \right)} \\\ & \Rightarrow \omega =\dfrac{v}{R}\sqrt{2\times 2{{\sin }^{2}}\dfrac{\theta }{2}} \\\ \end{aligned}
Therefore we can write that,
ω=2vsin⁡θ2R\omega =\dfrac{2v\sin \dfrac{\theta }{2}}{R}
Therefore the angular velocity has been calculated.

The correct answer has been mentioned as option A.

Note: Angular velocity is defined as the measure of the speed of an object revolves or rotates relative to another point. That is how fast the angular position or arrangement of an object varies with time. There are two kinds of angular velocity. They are the orbital angular velocity and spin angular velocity. Angular velocity is a vector quantity which is having a magnitude as well as direction.