Question

The angular momentum of a particle relative to the origin varies with time (\(t\)) as \(\vec{L} = (4\hat{i} + \alpha t^2 \hat{j})\, \mathrm{kg \cdot m}^2/\mathrm{s}\), where \(\alpha = 1\, \mathrm{kg \cdot m}^2/\mathrm{s}^3\). The angle between \(\vec{L}\) and the torque acting on the particle becomes \(45^\circ\) after a time of ............ s.
 

Hint:

Torque is the time derivative of angular momentum. The angle between them depends on how their vector components vary with time.

Answer 1

Correct Answer: 1.9

Step 1: Express torque. 
\[ \vec{\tau} = \frac{d\vec{L}}{dt} = (0\hat{i} + 2\alpha t \hat{j}) \]

Step 2: Compute the dot product. 
\[ \vec{L} \cdot \vec{\tau} = (4\hat{i} + \alpha t^2 \hat{j}) \cdot (0\hat{i} + 2\alpha t \hat{j}) = 2\alpha^2 t^3 \] \[ |\vec{L}| = \sqrt{(4)^2 + (\alpha t^2)^2} = \sqrt{16 + \alpha^2 t^4} \] \[ |\vec{\tau}| = 2\alpha t \]

Step 3: Use cosine relation. 
\[ \cos\theta = \frac{\vec{L}\cdot\vec{\tau}}{|\vec{L}||\vec{\tau}|} = \frac{2\alpha^2 t^3}{(2\alpha t)\sqrt{16 + \alpha^2 t^4}} = \frac{\alpha t^2}{\sqrt{16 + \alpha^2 t^4}} \] For \(\theta = 45^\circ,\ \cos\theta = \frac{1}{\sqrt{2}}\): \[ \frac{t^2}{\sqrt{16 + t^4}} = \frac{1}{\sqrt{2}} \Rightarrow t^4 = 16 \Rightarrow t = 2\, \text{s} \]

Step 4: Conclusion. 
Hence, the required time \(t = 2\, \text{s}\).