5.6 Newton’s Second Law in Rotational Form

Before the Fall 2024 course update, this topic was part of one big Rotation unit (old Unit 5). That unit is now split in two: Unit 5 covers spinning motion, torque and rotational inertia, and Unit 6 covers rotational energy, angular momentum and rolling. Older videos and practice labeled "Unit 5: Rotation" still fit, but they may mix in Unit 6 ideas.

For a rigid object turning about a fixed axis, $$\alpha = \frac{\tau_{\text{net}}}{I} \quad\text{or}\quad \tau_{\text{net}} = I\alpha$$ The angular acceleration is in the same direction as the net torque. It's proportional to the net torque and inversely proportional to the rotational inertia.

This is F = ma's twin: torque replaces force, I replaces mass, and α replaces a. Double the net torque and α doubles. Double I and α halves.

Take the torques and I about the same axis. For a fixed axle, that's the axle.

Real systems often have parts that move in straight lines and parts that rotate. You treat them separately: write ΣF = ma for each block and Στ = Iα for each rotating object. Then connect them.

For a string that wraps around a pulley without slipping, the block's acceleration equals the rim's tangential acceleration: $a = R\alpha$. That constraint is what lets you solve the equations together.

A pulley with mass needs a net torque to speed up. So the string tensions on its two sides are different. Only for a massless pulley are they equal.

Follow these steps for any system with blocks, strings and pulleys:

• Draw a separate diagram for each object: a free-body diagram for each block and a force diagram for each rotating object.
• Choose positive directions that agree. If the block moving down makes the pulley turn clockwise, call down positive for the block and clockwise positive for the pulley.
• Write ΣF = ma for each block and Στ = Iα for each pulley or wheel.
• Add the constraint a = Rα.
• Solve, then check the limits: if the pulley's mass goes to zero, your answer should match the massless-pulley result.

Wrap a string around a pulley's hub and hang different masses from it. For each mass, time how long the mass takes to fall a measured distance from rest. Then use kinematics to get a, and a = Rα to get α. The tension is T = m(g − a), so the torque is τ = TR.

Plot τ on the vertical axis against α. Newton's second law for rotation says the points should lie on a straight line with slope I. If the axle has a little friction, the line won't pass through the origin. The intercept tells you the friction torque you have to overcome before the pulley turns.

Moving a pivot changes two things at once: the rotational inertia and the torque from gravity. For example, a rod pivoted at its end and released from horizontal has α = 3g/(2L). Gravity acts at the center, at lever arm L/2, and I = ⅓ML².

Moving mass outward raises I. For the same torque, that makes α smaller. This is the reasoning behind many "increase, decrease or stay the same" questions.