Question 140: 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 60
o
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 60 o around the centre of the circle?

Choose the correct answer: