Question

A particle moves in a particle of radius 0.5 m at a speed that uniformly increases. Find the angular acceleration of particle if its speed changes from 2.0m//s to 4.0 m//s in 4.0s

Answer 1

Thus the value of angular acceleration is 1 rad/s^2

Explanation:

Given data:

Radius of circle = 0.5 m

Solution:

Tangential acceleration "at" = dv / dt

at = 4.0 - 2.0 / 4.0

at = 0.5 m/s^2

The angular acceleration is α=at / r

α = 0.5 / 0.5 =1 rad/s^2  

Thus the value of angular acceleration is 1 rad/s^2

Answer 2

Answer:

A body moves in a circle of radius 0.5m at a speed that increases uniformly. if the speed changes from 2m/s to 4m/s in 4s, its angular acceleration is $1rad/s^{2}$

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 $r=0.5m$

change in speed $v_{2} -v_{1}=4-2=2m/s$

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

⇒ $w_{2} -w_{1} =2/0.5=4rad/sec$

the angular acceleration is $4/4=1rad/s^{2}$