Question

inetial angular speed of a particle is 2 rad s-1 and contant angular acceleration is 3 rad rad s-2 then after 4 s it's angular displacement is .... rad​

Answer 1

Answer:

θ=ωίΤ+1/2αΤ^2

θ=(2)(4)+1/2(3)(4)^2

θ=8+24= 32 rad/s

Answer 2

Given :

• Initial angular speed (ω) = 2 rad/sec
• Angular acceleration (α) = 3 rad/sec
• Time Interval (t) = 4 seconds

To Find :

• Angular displacement (θ) = ?

Solution :

• DEFINITION :

◗ Angular speed (ω) : Rate of change of angular displacement of an object performing the circular motion is called as angular speed.

◗ Angular displacement (θ) : The angle traced out by the radius vector at the centre of the circular path in the given time interval is called as angular displacement.

◗Angular Acceleration (α) : The rate of change angular velocity of an object performing circular motion is called as angular acceleration.

• According to the Question Now :

↠ θ = ωt + ½ at²

↠ θ = 2 × 4 + ½ × 3 × (4)²

↠ θ = 8 + ½ × 3 × 16

↠ θ = 8 + 3 × 8

↠ θ = 8 + 24

↠ θ = 32 rad

∴ The angular displacement of the given particle is 32 rad.