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Question: An object of mass m has angular acceleration $\alpha=0.2 \text{rad s}^{-2}$. What is the angular dis...

An object of mass m has angular acceleration α=0.2rad s−2\alpha=0.2 \text{rad s}^{-2}. What is the angular displacement covered by the object after 3 second? (Assume that the object started with angle zero with zero angular velocity).

Answer

0.9 rad

Explanation

Solution

The problem asks for the angular displacement of an object undergoing constant angular acceleration.

Given:

  • Angular acceleration, α=0.2 rad s−2\alpha = 0.2 \text{ rad s}^{-2}
  • Time, t=3 st = 3 \text{ s}
  • Initial angular position, θ0=0 rad\theta_0 = 0 \text{ rad} (since it started with angle zero)
  • Initial angular velocity, ω0=0 rad s−1\omega_0 = 0 \text{ rad s}^{-1} (since it started with zero angular velocity)

We can use the kinematic equation for angular displacement under constant angular acceleration:

θ=θ0+ω0t+12αt2\theta = \theta_0 + \omega_0 t + \frac{1}{2} \alpha t^2

Substitute the given values into the equation:

θ=0+(0)(3)+12(0.2)(3)2=0+0+12(0.2)(9)=(0.1)(9)=0.9 rad\theta = 0 + (0)(3) + \frac{1}{2} (0.2) (3)^2 = 0 + 0 + \frac{1}{2} (0.2) (9) = (0.1) (9) = 0.9 \text{ rad}

Thus, the angular displacement covered by the object after 3 seconds is 0.9 rad.