Question 131: Work, Energy and Power & Circular Motion

Question

A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o around the centre of the circle?

Accepted Answer

$$\left| {\Delta \overrightarrow v } ight| = \sqrt {v_1^2 + v_2^2 + 2{v_1}{v_2}\cos \left( {\pi  - 	heta } ight)} $$
$$ = 2v\sin {	heta  \over 2}$$        since $$\left[ {\left| {\overline v {}_1} ight| = \left| {{{\overline v }_2}} ight|} ight]$$
$$ = \left( {2 	imes 10} ight) 	imes \sin \left( {{{30}^o}} ight)$$
= 10 m/s

Answer

zero

Answer

10 m/s

Answer

$$10\sqrt 2 m/s$$

Answer

$$10\sqrt 3 m/s$$

Work, Energy and Power & Circular Motion

A particle is moving along a circular path with a constant speed of 10 ms–1. What is the magnitude of the change in velocity of the particle, when it moves through an angle of 60o around the centre of the circle?

Choose the correct answer: