Question

A body is describing a circular motion with a speed v,when it describes an angle thita at the center ,the change in velocity delta v is

Answer 1

Given:

• Particle describes circular motion.
• It describes an angle theta ∅ at centre.
• Speed of particle is v

To find:

• Change in velocity (Δv)

Solution:

• Let speed after a certain time be v₂ and initial velocity be v₁ but both will be v since it is given.
• Change in velocity = v₂-v₁
• v₂-v₁ = √(v₂² + v₁²  -2v₁v₂ cos ∅)
• v₂-v₁ = √(v²+v² - 2v²cos∅)
• v₂-v₁ = √(2v²{1-cos∅}) .... (∵ Taking 2v² common)
• Using half angle formula of trigonometry,
• v₂-v₁ = √(2v²*2sin² {∅/2} )  ........ (∵ 1-cos∅ = 2sin²{∅/2})
•        = $\sqrt{ 4v^2 sin (\frac{\theta}{2} )}$  
• Δv = 2v sin (∅/2)  

Answer:

The change in velocity Δv = 2v sin(∅/2).