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Rotational dynamics: for a rigid body with moment of inertia I and angular acceleration α, identify the correct expression for the applied torque.

Difficulty: Easy

Correct Answer: I α

Explanation:


Introduction / Context:
The fundamental relation between torque and angular acceleration in rotational dynamics mirrors Newton’s second law for translation. Correctly applying tau = I * alpha is essential in analyzing gears, flywheels, rotors, and any accelerating rotating machinery.


Given Data / Assumptions:

  • Rigid body rotating about a fixed axis.
  • Moment of inertia about that axis is I (kg·m^2).
  • Angular acceleration is alpha (rad/s^2).


Concept / Approach:
Analogous to F = m * a, rotational motion follows tau = I * alpha, where tau is the net torque about the axis. Dimensional consistency: N·m on LHS; on RHS, (kg·m^2) * (rad/s^2). Since rad is dimensionless, kg·m^2/s^2 equals N·m, confirming the relation.


Step-by-Step Solution:

Start from rotational form of Newton’s second law: sum(tau) = I * alpha.Identify I and alpha from the problem statement.Conclude tau = I * alpha (no squares, no inverses).


Verification / Alternative check:

Check units: I (kg·m^2) * alpha (1/s^2) = kg·m^2/s^2 = N·m → matches torque.


Why Other Options Are Wrong:

I2 α and I2/α: squaring I is dimensionally incorrect for torque.I/α: inversion produces dimensions of kg·m^2·s^2, not torque.I α1/2: taking a square root of alpha lacks physical basis and breaks dimensionality.


Common Pitfalls:

Confusing linear and angular forms (F = m * a vs tau = I * alpha).Forgetting that radians are dimensionless, which keeps units consistent.


Final Answer:

I α
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