The problem provided in the image involves a rotating disk, and you are asked to calculate several aspects related to its rotational motion. Let's break down the problem and solve each part.

Problem Breakdown:

Given:

• The torque ($\tau(t)$) applied to the disk is given by the equation: $\tau(t) = 5 + 4t^2 \, \text{Nm}$
• The mass of the disk ($m$) is 10 kg.
• The disk can be modeled as a uniform solid disk.
• The time at which the disk is analyzed is $t = 3$ seconds.
• The radius of the disk is $R = 0.6$ m.

Part (a) Determine the angular acceleration at $t = 3$ seconds.

1. Moment of Inertia for a Solid Disk: The moment of inertia $I$ for a solid disk rotating about its central axis is given by: $I = \frac{1}{2} m R^2$ Substituting the given values: $I = \frac{1}{2} \times 10 \, \text{kg} \times (0.6 \, \text{m})^2 = 1.8 \, \text{kg} \cdot \text{m}^2$

2. Angular Acceleration: The angular acceleration $\alpha(t)$ can be found using the relationship between torque and angular acceleration: $\tau(t) = I \alpha(t)$ Therefore: $\alpha(t) = \frac{\tau(t)}{I}$ At $t = 3$ seconds, the torque is: $\tau(3) = 5 + 4(3)^2 = 5 + 36 = 41 \, \text{Nm}$ Substituting the values: $\alpha(3) = \frac{41 \, \text{Nm}}{1.8 \, \text{kg} \cdot \text{m}^2} \approx 22.78 \, \text{rad/s}^2$

Part (b) Determine the angular velocity after 3 seconds.

1. Initial Angular Velocity: We are not given an explicit initial angular velocity, so we assume it starts from rest, i.e., $\omega(0) = 0$.

2. Angular Velocity: Angular acceleration is the derivative of angular velocity: $\frac{d\omega}{dt} = \alpha(t)$ Therefore, the angular velocity at time $t = 3$ seconds can be found by integrating the angular acceleration: $\omega(t) = \int_0^t \alpha(t) \, dt$ Given $\alpha(t) = \frac{\tau(t)}{I} = \frac{5 + 4t^2}{1.8}$, we integrate from 0 to 3 seconds: $\omega(3) = \int_0^3 \frac{5 + 4t^2}{1.8} \, dt$ Let's compute this integral.

   $\omega(3) = \frac{1}{1.8} \left[ 5t + \frac{4t^3}{3} \right]_0^3$ $\omega(3) = \frac{1}{1.8} \left[ 5(3) + \frac{4(3)^3}{3} \right]$ [ \omega(3) = \frac{1}{1.8} \left[ 15 + \frac{4(27)}{3} \right] = \frac{1}{1.8} \left[ 15 + 36 \right] = \frac{1}{1.8} \times 51 \approx 28.33 , \text{rad/s} ]

Part (c) What is the direction of the torque at $t = 3$ seconds?

Since torque ($\tau(t)$) is a scalar equation and positive in this case, it acts in the direction of the applied force, which is counterclockwise if we assume the positive direction follows the right-hand rule.

Summary of Answers:

• (a) Angular acceleration at $t = 3$ seconds: $\alpha(3) \approx 22.78 \, \text{rad/s}^2$
• (b) Angular velocity after 3 seconds: $\omega(3) \approx 28.33 \, \text{rad/s}$
• (c) The torque direction is counterclockwise.

Would you like a more detailed explanation, or do you have any questions?

Follow-up Questions:

1. What is the physical significance of angular acceleration in rotational motion?
2. How would the moment of inertia change if the mass of the disk were doubled?
3. What would happen to the angular velocity if the initial angular velocity was not zero?
4. How does torque relate to rotational kinetic energy?
5. What role does the radius of the disk play in determining its moment of inertia?

Tip:

When dealing with rotational dynamics, always remember to check the units of each quantity (e.g., torque in Nm, angular velocity in rad/s) to ensure consistency in your calculations.