Question

The speed of a particle revolving in a
circle of radius 0.23 m, changes
uniformly from 1m/s to 2m/s in 2s.
The angular acceleration of the particle
(in rad/s^2 ) is​

Answer 1

Answer:

2.174 rad/ s²

angular acceleration is rate of change of angular velocity.

$\alpha = \frac{Δω}{Δt} = \frac{Δv}{rΔt}$

refer pic

Answer 2

Answer:

The speed of a particle revolving in a  circle of radius 0.23 m, changes  uniformly from 1m/s to 2m/s in 2s. The angular acceleration of the particle

(in rad/s^2 ) is​ $2.17rad/s^{2}$

Explanation:

• The angular acceleration is the change in angular velocity divided by time to change

        $\alpha =\frac{w_{2} -w_{1} }{t_{2} -t_{1} }$

• In a circular motion, the linear velocity (V) and angular velocity (w) are related as,

        $V=rw$

        where r-radius of the circular path

From the question,

the radius of the circular path $r=0.23m$

change in speed $v_{2} -v_{1}=2-1=1m/s$

so the change in angular velocity $w_{2} -w_{1} =\frac{v_{2} -v_{1} }{r}$

⇒ $w_{2} -w_{1} =1/0.23=4.3rad/sec$

the time taken $t=2s$

the angular acceleration is $4.3/2=2.17rad/s^{2}$