Let's address the problem step-by-step.
### Given:
- Rotational inertia [tex]\[ I = 2.4 \times 10^{-2} \, \text{kg·m}^2 \][/tex]
- Radius [tex]\[ r = 0.11 \, \text{m} \][/tex]
- Force as a function of time [tex]\[ F(t) = 0.60 t + 0.30 t^2 \][/tex]
- Initial conditions: the pulley is at rest, which means initial angular speed [tex]\[ \omega_0 = 0 \][/tex]
- Time [tex]\[ t = 4.9 \, \text{s} \][/tex]
### (a) Angular acceleration
1. Calculate the force at time [tex]\[ t = 4.9 \, \text{s} \][/tex]:
[tex]\[ F(4.9) = 0.60 \cdot 4.9 + 0.30 \cdot (4.9)^2 \][/tex]
2. Compute the force:
[tex]\[ F(4.9) = 0.60 \cdot 4.9 + 0.30 \cdot 24.01 \][/tex]
[tex]\[ F(4.9) = 2.94 + 7.203 \][/tex]
[tex]\[ F(4.9) = 10.143 \, \text{N} \][/tex]
3. Torque [tex]\[ \tau \][/tex] is given by force applied tangentially times the radius:
[tex]\[ \tau = F(4.9) \cdot r \][/tex]
[tex]\[ \tau = 10.143 \cdot 0.11 \][/tex]
[tex]\[ \tau = 1.11573 \, \text{Nm} \][/tex]
4. Angular acceleration [tex]\[ \alpha \][/tex] can be calculated using the formula:
[tex]\[ \alpha = \frac{\tau}{I} \][/tex]
[tex]\[ \alpha = \frac{1.11573}{2.4 \times 10^{-2}} \][/tex]
[tex]\[ \alpha = 46.48875 \, \text{rad/s}^2 \][/tex]
So, the angular acceleration at [tex]\( t = 4.9 \)[/tex] s is:
[tex]\[ \alpha = 46.48875 \, \text{rad/s}^2 \][/tex]
### (b) Angular speed
5. To find the angular speed [tex]\[ \omega \][/tex], we need to integrate the angular acceleration over time.
The angular speed is given by integrating the angular acceleration from 0 to 4.9 seconds:
[tex]\[ \omega = \int_0^{4.9} \alpha(t) \, dt \][/tex]
Given that the angular acceleration is derived from the force function, we have:
[tex]\[ \alpha(t) = \frac{r (0.60 t + 0.30 t^2)}{I} \][/tex]
6. Integrate the function over the given interval:
[tex]\[ \omega = \int_0^{4.9} \frac{r (0.60 t + 0.30 t^2)}{I} \, dt \][/tex]
[tex]\[ \omega = \frac{r}{I} \int_0^{4.9} (0.60 t + 0.30 t^2) \, dt \][/tex]
7. Calculate the integral:
[tex]\[ \int_0^{4.9} (0.60 t + 0.30 t^2) \, dt = 0.60 \int_0^{4.9} t \, dt + 0.30 \int_0^{4.9} t^2 \, dt \][/tex]
[tex]\[ = 0.60 \left[ \frac{t^2}{2} \right]_0^{4.9} + 0.30 \left[ \frac{t^3}{3} \right]_0^{4.9} \][/tex]
[tex]\[ = 0.60 \cdot \frac{(4.9)^2}{2} + 0.30 \cdot \frac{(4.9)^3}{3} \][/tex]
[tex]\[ = 0.60 \cdot 12.005 + 0.30 \cdot 39.207 \][/tex]
[tex]\[ = 7.203 + 11.7621 \][/tex]
[tex]\[ = 18.9651 \][/tex]
8. Multiply by the factor [tex]\[ \frac{r}{I} \][/tex]:
[tex]\[ \omega = \frac{0.11}{2.4 \times 10^{-2}} \cdot 18.9651 \][/tex]
[tex]\[ \omega = 4.583333 \cdot 18.9651 \][/tex]
[tex]\[ \omega = 70.429333 \, \text{rad/s} \][/tex]
So, the angular speed at [tex]\( t = 4.9 \)[/tex] s is:
[tex]\[ \omega = 70.429333 \, \text{rad/s} \][/tex]
### Summary:
(a) The angular acceleration [tex]\(\alpha\)[/tex] at [tex]\( t = 4.9 \)[/tex] s is [tex]\( 46.48875 \, \text{rad/s}^2 \)[/tex].
(b) The angular speed [tex]\(\omega\)[/tex] at [tex]\( t = 4.9 \)[/tex] s is [tex]\( 70.429333 \, \text{rad/s} \)[/tex].
