Question

When a torque of 32.0 Nm is applied to a certain wheel, the wheel acquires an angular acceleration of 25.0 units. What is the rotational inertia of the wheel?

• A. \(1.25 \text{ kg}\cdot\text{m}^2\)
• B. \(1.28 \text{ kg}\cdot\text{m}^2\)
• C. \(2.28 \text{ kg}\cdot\text{m}^2\)
• D. \(2.25 \text{ kg}\cdot\text{m}^2\)

Hint:

To divide any number by 25 quickly in your head, multiply the number by 4 and shift the decimal point two places to the left: \[ 32 \times 4 = 128 \quad \Rightarrow \quad 1.28 \]

Answer 1

The Correct Option is B

Concept: Newton's Second Law for rotational motion states that the net external torque (\(\tau\)) applied to a rigid body is directly proportional to the resulting angular acceleration (\(\alpha\)). The constant of proportionality is the rotational inertia (moment of inertia, \(I\)) of the object about that axis of rotation: \[ \tau = I\alpha \]

Step 1: Identifying the given parameters.
From the problem description, we have:
• Applied torque, \(\tau = 32.0 \text{ N}\cdot\text{m}\)
• Angular acceleration, \(\alpha = 25.0 \text{ rad/s}^2\) (units)

Step 2: Solving for the rotational inertia \(I\).
Rearranging the rotational analogue of Newton's second law equation to isolate \(I\): \[ I = \frac{\tau}{\alpha} \] Substitute the values into this equation: \[ I = \frac{32.0}{25.0} \]

Step 3: Calculating the final numerical value.
To easily evaluate this fraction, we can multiply both the numerator and the denominator by 4: \[ I = \frac{32 \times 4}{25 \times 4} = \frac{128}{100} = 1.28 \text{ kg}\cdot\text{m}^2 \] This matches option (2).