Certainly! Let's break down the problem step by step:
### Part (a): Calculate Angular Momentum
We are given:
- Torque ([tex]\( \tau \)[/tex]) = 0.12 [tex]\( \text{N} \cdot \text{m} \)[/tex]
- Time ([tex]\( t \)[/tex]) = 0.65 [tex]\( \text{s} \)[/tex]
Angular momentum ([tex]\( L \)[/tex]) is given by the product of torque and time:
[tex]\[ L = \tau \cdot t \][/tex]
By substituting the given values:
[tex]\[ L = 0.12 \, \text{N} \cdot \text{m} \times 0.65 \, \text{s} \][/tex]
So, the angular momentum [tex]\( L \)[/tex] is:
[tex]\[ L = 0.078 \, \text{N} \cdot \text{m} \cdot \text{s} \][/tex]
### Part (b): Calculate Angular Speed
We are additionally given:
- Moment of Inertia ([tex]\( I \)[/tex]) = [tex]\( 2.5 \times 10^{-3} \, \text{kg} \cdot \text{m}^2 \)[/tex]
First, find the angular acceleration ([tex]\( \alpha \)[/tex]). Angular acceleration is given by:
[tex]\[ \alpha = \frac{\tau}{I} \][/tex]
By substituting the given values:
[tex]\[ \alpha = \frac{0.12 \, \text{N} \cdot \text{m}}{2.5 \times 10^{-3} \, \text{kg} \cdot \text{m}^2} \][/tex]
So, the angular acceleration [tex]\( \alpha \)[/tex] is:
[tex]\[ \alpha = 48.0 \, \text{rad/s}^2 \][/tex]
Now, to find the angular speed ([tex]\( \omega \)[/tex]) after time [tex]\( t \)[/tex]:
[tex]\[ \omega = \alpha \cdot t \][/tex]
By substituting the values of [tex]\( \alpha \)[/tex] and [tex]\( t \)[/tex]:
[tex]\[ \omega = 48.0 \, \text{rad/s}^2 \times 0.65 \, \text{s} \][/tex]
So, the angular speed [tex]\( \omega \)[/tex] is:
[tex]\[ \omega = 31.2 \, \text{rad/s} \][/tex]
### Summary:
- Angular Momentum: [tex]\( 0.078 \, \text{N} \cdot \text{m} \cdot \text{s} \)[/tex]
- Angular Speed: [tex]\( 31.2 \, \text{rad/s} \)[/tex]
