Question

A particle of mass 'm' is moving in time 't' on a trajectory given by \[ \vec{r} = 10 \alpha t^2 \hat{i} + 5 \beta (t - 5) \hat{j} \] Where $\alpha$ and $\beta$ are dimensional constants. The angular momentum of the particle becomes the same as it was for $t=0$ at time $t = $________ seconds.

Hint:

Always simplify the cross product into a scalar function of time. The angular momentum returns to zero when the position vector and velocity vector become parallel.

Answer 1

Correct Answer: 10

Step 1: Understanding the Concept:
Angular momentum is defined as the cross product of the position vector and the linear momentum. We need to find the time when this value equals its initial value at \(t=0\).
Step 2: Key Formula or Approach:
1. Velocity: \(\vec{v} = \frac{d\vec{r}}{dt}\).
2. Angular momentum: \(\vec{L} = m(\vec{r} \times \vec{v})\).
Step 3: Detailed Explanation:
Velocity \(\vec{v}\):
\[ \vec{v} = \frac{d}{dt} [10 \alpha t^2 \hat{i} + (5 \beta t - 25 \beta) \hat{j}] = 20 \alpha t \hat{i} + 5 \beta \hat{j} \]
Now, calculate \(\vec{r} \times \vec{v}\):
\[ \vec{r} \times \vec{v} = [ (10 \alpha t^2 \hat{i} + (5 \beta t - 25 \beta) \hat{j}) \times (20 \alpha t \hat{i} + 5 \beta \hat{j}) ] \]
\[ = [ (10 \alpha t^2)(5 \beta) \hat{k} - (5 \beta t - 25 \beta)(20 \alpha t) \hat{k} ] \]
\[ = [ 50 \alpha \beta t^2 - (100 \alpha \beta t^2 - 500 \alpha \beta t) ] \hat{k} \]
\[ = [ 500 \alpha \beta t - 50 \alpha \beta t^2 ] \hat{k} \]
At \(t = 0\), \(\vec{L} = 0\).
For \(\vec{L}(t) = 0\) again:
\[ 500 \alpha \beta t - 50 \alpha \beta t^2 = 0 \implies 50 \alpha \beta t (10 - t) = 0 \]
Since \(t \neq 0\):
\[ 10 - t = 0 \implies t = 10 \text{ s} \]
Step 4: Final Answer:
The angular momentum is the same at \(t = 10\) seconds.