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

To find the angular acceleration of a particle at a given time, we first need to understand the relationship between angular position, angular velocity, and angular acceleration.

Angular acceleration \(\alpha\) is the derivative of angular velocity \(\omega\) with respect to time. Angular velocity \(\omega\) is the derivative of angular position \(\theta\) with respect to time.

Given the angular position as a function of time: \(\theta = \frac{t^4}{4} + t^2\)

First, find the angular velocity \(\omega\) by differentiating \(\theta\) with respect to \(t\):

\(\omega = \frac{d\theta}{dt} = \frac{d}{dt} \left(\frac{t^4}{4} + t^2\right)\)

\(= \left(\frac{1}{4}\right) \cdot 4t^3 + 2t\)

\(= t^3 + 2t\)

Then, find the angular acceleration \(\alpha\) by differentiating \(\omega\) with respect to \(t\):

\(\alpha = \frac{d\omega}{dt} = \frac{d}{dt} (t^3 + 2t)\)

\(= 3t^2 + 2\)

Substitute \(t = 1\, \text{s}\) to find the angular acceleration at that time:

\(\alpha = 3(1)^2 + 2\)

\(= 3 \cdot 1 + 2\)

\(= 5\, \text{rad/s}^2\)

Hence, the angular acceleration at \(t = 1\, \text{s}\) is \(5\, \text{rad/s}^2\), which matches the given correct answer option.