Question

16. A particle of mass 5 g is moving in a circle of radius
0.5 m with an angular velocity of 6 rad/s. Find :
(1) the change in linear velocity in half a revolution,
sn the magnitude of the acceleration of the particle.
Ans. (1) 6 m/s, (m) 18 m/s​

Answer 1

Change in velocity will be v-(-v)

Explanation:since as the body moves in circle its direction is changing but speed is remaining constant

Answer 2

Answer:

Change in linear velocity is 6 m/s and the magnitude of the acceleration of the particle is 18 $m/s^{2}$

Explanation:

Since mass of the particle(m) =5 g.

Radius of circle (R) 0.5 m.

Angular velocity (ω) 6 rad/s.

We know the relation between linear velocity,angular velocity and radius of the circle.

V= ωR

V=6 x 0.5=3 m/s.

The change in linear velocity in half a revolution of circle = V-(-V)=2V

                                                                                               =2 x 3=6 m/s.

We know that the centripetal acceleration is given by $a_{c} =\dfrac{V^{2} }{R} = \dfrac{9}{0.5}=18\ m/s^{2}$

Change in linear velocity is 6 m/s and the magnitude of the acceleration of the particle is 18 $m/s^{2}$