Question

A particle is rotating in a circular path and at any instant its motion can be described as} \[ \theta=\frac{5t^4}{40}-\frac{t^3}{3}. \] The angular acceleration of the particle after \(10\) seconds is ____ rad/s\(^2\).

• A. \(150\)
• B. \(120\)
• C. \(130\)
• D. \(170\)

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Angular acceleration (\(\alpha\)) is the rate of change of angular velocity (\(\omega\)) with respect to time, and angular velocity is the rate of change of angular displacement (\(\theta\)) with respect to time.
Mathematically, \(\omega = \frac{d\theta}{dt}\) and \(\alpha = \frac{d\omega}{dt} = \frac{d^2\theta}{dt^2}\).
Step 2: Key Formula or Approach:
1. \(\omega = \frac{d\theta}{dt}\)
2. \(\alpha = \frac{d\omega}{dt}\)
Step 3: Detailed Explanation:
Given angular displacement:
\[ \theta = \frac{5t^4}{40} - \frac{t^3}{3} = \frac{t^4}{8} - \frac{t^3}{3} \]
Differentiate \(\theta\) with respect to time \(t\) to find angular velocity \(\omega\):
\[ \omega = \frac{d}{dt} \left( \frac{t^4}{8} - \frac{t^3}{3} \right) = \frac{4t^3}{8} - \frac{3t^2}{3} = \frac{t^3}{2} - t^2 \]
Differentiate \(\omega\) with respect to time \(t\) to find angular acceleration \(\alpha\):
\[ \alpha = \frac{d}{dt} \left( \frac{t^3}{2} - t^2 \right) = \frac{3t^2}{2} - 2t \]
Now, calculate the value of \(\alpha\) at \(t = 10 \text{ s}\):
\[ \alpha_{t=10} = \frac{3(10)^2}{2} - 2(10) \]
\[ \alpha_{t=10} = \frac{300}{2} - 20 = 150 - 20 = 130 \text{ rad/s}^2 \]
Step 4: Final Answer:
The angular acceleration after 10 seconds is 130 \(\text{rad/s}^2\).