Question

An object is moving on a circular trajectory with uniform speed "v". It moves by an angle $\theta$ on the circular path.

Prove :

The change in velocity of the object is :

$\boxed{ \sf\Delta v = 2v \sin \bigg( \dfrac{ \theta}{2} \bigg ) }$
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Answer 1

$\underline{\underline{\sf{\maltese\:\:Question}}}$

• An object is moving on a circular trajectory with uniform speed $\sf{v}$. It moves by an angle $\theta$ on the circular path.  Prove The change in velocity of the object is : $\sf{\Delta v=2 v \sin \left(\theta/2\right)}$

$\underline{\underline{\sf{\maltese\:\:Given}}}$

• An object is moving on a circular trajectory with uniform speed $\sf{v}$
• It moves by an angle $\theta$ on the circular path

$\underline{\underline{\sf{\maltese\:\:To\:Prove}}}$

• The change in velocity of the object is :

$\sf{\Delta v=2 v \sin \left(\dfrac{\theta}{2}\right)}$

$\underline{\underline{\sf{\maltese\:\:Answer}}}$

• When an object moves on a circular trajectory with uniform speed, the velocity constantly changes due to the constant change in direction of motion of the object. By speed we consider the magnitude of the object's velocity

The situation in the question is represented in the diagram below.

• (Kindly Refer to Attachment)

Let point A be the initial position of the object and point B be the position of the object at angle $\theta$ away.

• Velocity at point A can be written as, $\sf{\overrightarrow{\sf{v}_{\sf{i}}}=v \hat{j}}$

• Velocity at point B can be written as, $\sf{\overrightarrow{\sf{v}_{\sf{f}}}=-v \sin \theta \hat{i}+v \cos \theta \hat{j}}$

• The magnitude of velocity i.e. speed is always $\sf{v}$

The change in velocity is given by, $\sf{\overrightarrow{\sf{v}_{f}}-\overrightarrow{\sf{v}_{i}}=-v \sin \theta \hat{i}+(v \cos \theta-v) \hat{j}}}$

The magnitude of change in velocity is given by,

$\to\sf{\Delta v=\sqrt{(-v \sin \theta)^{2}+(v \cos \theta-v)^{2}}}$

$\to\sf{\Delta v=\sqrt{v^{2} \sin ^{2} \theta+v^{2} \cos ^{2} \theta-2 v^{2} \cos \theta+v^{2}}}$

$\to\sf{\Delta v=\sqrt{v^{2}\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+v^{2}-2 v^{2} \cos \theta}}$

$\to\sf{\Delta v=\sqrt{v^{2}\left(1\right)+v^{2}-2 v^{2} \cos \theta}}$

$\to\sf{\Delta v=\sqrt{v^{2}+v^{2}-2 v^{2} \cos \theta}}$

$\to\sf{\Delta v=\sqrt{2 v^{2}(1-\cos \theta)}}$

$\to\sf{\Delta v=\sqrt{2 v^{2} \times 2 \sin ^{2} (\theta/2)}}$

$\to\sf{\Delta v=2 v \sin \left(\dfrac{\theta}{2}\right)}$

• $\bf{Hence\:Proved\:!!!}$

We have proved that for an object moving with uniform speed $\sf{v}$ on a circular trajectory, the change in velocity in moving through angle $\theta$ is,

$\boxed{\sf{\Delta v=2 v \sin \left(\dfrac{\theta}{2}\right)}}$

Answer 2

Answer:

$\boxed{\sf \Delta v =2v\sin \Bigg(\dfrac{\theta}{2}\Bigg)}$

Explanation:

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Let the velocities at positions A and B be v₁ and v₂ respectively. We need to find the change in velocity i.e. Δv. Rearranging the v₁, v₂ & Δv as in fig. 2. So, from triangle law of vectors we can say that v₁ + Δv = v₂, Therefore,

⇒ Δv = v₂ - v₁

And we also know that |v₂| = |v₁| =|v|,

Applying law of vector subtraction,

$\longrightarrow\sf \Delta v=\sqrt{\bigg(v_{1}\bigg)^{2}+\bigg(v_{2}\bigg)^{2}-2v_{1}v_{2}\cos\theta}$

Substituting the values,

$\longrightarrow\sf \Delta v=\sqrt{\bigg(v\bigg)^{2}+\bigg(v\bigg)^{2}-2\times v\times v \times \cos\theta}\\\\\\\\\longrightarrow\sf \Delta v=\sqrt{\bigg(2\;v^{2}\bigg)-2\;v^{2}\times \cos\theta}\\\\\\\\\longrightarrow\sf \Delta v = \sqrt{2v^{2}\bigg(1-\cos\theta\bigg)}\ \ \ \ \because\Bigg[\bigg(1-\cos\theta\bigg)=2\sin^{2}\bigg(\dfrac{\theta}{2}\bigg)\Bigg]\\\\\\\\\longrightarrow\sf \Delta v = \sqrt{2v^{2}\times2\sin^{2}\bigg(\dfrac{\theta}{2}\bigg)}$

$\longrightarrow\sf \Delta v =\sqrt{4\;v^{2}\sin^{2}\bigg(\dfrac{\theta}{2}\bigg)}\\\\\\\\\longrightarrow\sf \Delta v =2v\sin\bigg(\dfrac{\theta}{2}\bigg)\\\\\\\\\longrightarrow\large{\underline{\boxed{\sf \Delta v =2v\sin\bigg(\dfrac{\theta}{2}\bigg) }}}$

Hence Proved!

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