Question

If angular position of a particle is given by $\theta = \frac{t^4}{4} + t^2$. Find angular acceleration at $t = 1s$:

• A. $6 \text{ rad/s}^2$
• B. $5 \text{ rad/s}^2$
• C. $10 \text{ rad/s}^2$
• D. $6 \text{ rad/s}^2$

Answer 1

The Correct Option is B

Step 1: State the given expression for angular position.
The angular position $\theta$ as a function of time $t$ is:
$ \theta = \frac{t^4}{4} + t^2 $

Step 2: Find the angular velocity ($\omega$).
Angular velocity is the rate of change of angular position with respect to time.
$ \omega = \frac{d\theta}{dt} = \frac{d}{dt} \left( \frac{t^4}{4} + t^2 \right) $
$ \omega = t^3 + 2t $

Step 3: Find the angular acceleration ($\alpha$).
Angular acceleration is the rate of change of angular velocity with respect to time.
$ \alpha = \frac{d\omega}{dt} = \frac{d}{dt} (t^3 + 2t) $
$ \alpha = 3t^2 + 2 $

Step 4: Calculate angular acceleration at $t = 1 \text{ s}$.
Substitute $t = 1$ into the equation for $\alpha$:
$ \alpha(t=1) = 3(1)^2 + 2 $
$ \alpha = 5 \text{ rad/s}^2 $

Final Answer: The angular acceleration is $5 \text{ rad/s}^2$. Correct option is (2).