Newton’s First Law in Rotational Form (College Board AP® Physics 1: Algebra-Based)

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Physics

Newton’s first law in rotational form

The rotational analog of Newton’s first law states:

If the net torque exerted on a system is zero, the angular velocity of that system will remain constant

A constant angular velocity could also be an angular velocity of zero, i.e. when the system is not rotating
If the net torque acting on a system is zero, it is said to be in rotational equilibrium

Unbalanced torque

According to Newton’s second law, if the torques exerted on a rigid system are not balanced, the system’s angular velocity must be changing
Therefore, a net torque produces an angular acceleration
The direction of this angular acceleration depends on the direction of the net torque

Beam with an unbalanced torque

A diagram showing a lever with forces applied at three different points, labeled F1, F2, and F3, with their respective distances from the pivot indicated as r1, r2, and r3. Text boxes below show torque equations. — The diagram shows two different scenarios: net torque in the clockwise and counterclockwise directions. If there is a net torque in the clockwise or counterclockwise direction, the beam will also have an angular acceleration in that direction.

Worked Example

Three forces act on a wheel which is free to rotate about its center O, as shown in the diagram.

Diagram showing a circular disk with a central point O. Various force vectors (10 N, 12 N, 9 N) are applied at different points and angles. Dimensions are 10 cm and 25 cm.

Calculate the net torque about O and state whether the angular acceleration that is produced is clockwise or counterclockwise.

Answer:

Step 1: Analyze the scenario

The magnitude of the torque about the axis of rotation O is given by

$τ = F r \sin θ$

The 9 N force and the 10N force are both perpendicular to the radius and produce a torque in the counterclockwise direction
The perpendicular component of the 12 N force produces a torque in the clockwise direction

Diagram shows a circle with forces: 12 N at 30 degrees to the left, 10 N to the right, 9 N downward, and components 12sin30° downward. Radii are 10 cm and 25 cm.

Step 2: Calculate the total counterclockwise torque

The torque due to the 10 N force is

$τ = 10 \times 0.25 \times \sin 90 = 2.5 N \cdot m$

The torque due to the 9 N force is

$τ = 9 \times 0.25 \times \sin 90 = 2.25 N \cdot m$

Therefore, the total counterclockwise torque = 2.5 + 2.25 = 4.75 N m

Step 3: Calculate the total clockwise torque

The torque due to the 12 N force is:

$τ = 12 \times 0.1 \times \sin 30 = 0.6 N \cdot m$

Step 4: Determine the net torque and direction of angular acceleration

Net torque = total counterclockwise torque − total clockwise torque

Net torque = $\sum τ = 4.75 - 0.6 = 4.15 N \cdot m$ , counterclockwise

The net torque and angular acceleration both act in the counterclockwise direction

Examiner Tip

In the AP Physics 1 exam, you will not be expected to analyze rotation in multiple planes. When analyzing the torques exerted on a system, you only need to be able to consider whether they produce clockwise or counterclockwise motion

Diagram of a lever with a pivot point. It shows the effects of forces labeled clockwise and anti-clockwise on either side of the pivot.

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