Question

For mass 20 kg, torque \( \tau = 5 \, \text{Nm} \), find angular acceleration \( \alpha \).

Hint:

Angular acceleration is found using \( \alpha = \frac{\tau}{I} \), where \( I \) is the moment of inertia. For a point mass, \( I = mr^2 \).

Answer 1

Step 1: Understanding the relationship between torque and angular acceleration.
Torque \( \tau \) is related to angular acceleration \( \alpha \) by the equation: \[ \alpha = \frac{\tau}{I} \] where \( I \) is the moment of inertia. For a point mass, the moment of inertia is given by \( I = mr^2 \), where \( m \) is the mass and \( r \) is the radius or distance from the axis of rotation.
Step 2: Substituting the known values.
Given \( \tau = 5 \, \text{Nm} \) and mass \( m = 20 \, \text{kg} \), we need the radius \( r \) to calculate \( I \). If the radius is provided, we can calculate \( I \) and then use the equation for \( \alpha \).
Step 3: Conclusion.
Thus, the angular acceleration \( \alpha \) can be calculated using \( \alpha = \frac{\tau}{I} \), with \( I = mr^2 \) for a point mass or the appropriate moment of inertia formula for other shapes.

Answer 2

Step 1: Torque and angular acceleration are linked by τ = I × α, so α = τ ÷ I.

Step 2: For a point mass, the moment of inertia is I = m × r², where r is the distance from the rotation axis.

Step 3: Here mass m = 20 kg and torque τ = 5 N·m.

Step 4: Once the radius r is known, calculate I = mr², then find α = τ ÷ I.

Answer 3

Building up from Newton's second law for linear motion.
For a point mass moving in a circle of radius \( r \), the tangential force needed is \( F = ma = m(r\alpha) \), since tangential acceleration \( a = r\alpha \).
Torque is this force times the radius, \( \tau = Fr = m r^2 \alpha \), so \[ \alpha = \frac{\tau}{mr^2} \]
With mass \( m = 20 \, \text{kg} \) and torque \( \tau = 5 \, \text{N·m} \), \[ \alpha = \frac{5}{20 r^2} = \frac{0.25}{r^2} \] once the radius \( r \) is known.