Torque & Cross Products - Physics Q and A

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Visualization of a heavy metal door with a glowing force vector pushing it, producing a perpendicular torque vector along the hinge axis.

Force causes linear acceleration. Torque ( $\tau$ ) causes angular acceleration.

In AP Physics 1, Torque is often simplified to “Force times lever arm.” In AP Physics C, you must understand Torque in its true mathematical form: a Vector Cross Product. The direction you push, the distance from the pivot, and the 3D angle between them all dictate how effectively you can cause an object to spin.

1. Torque as a Cross Product

Torque ( $\vec{\tau}$ ) is defined as the cross product of the position vector ( $\vec{r}$ ) and the force vector ( $\vec{F}$ ). The resulting torque vector points perpendicular to both $\vec{r}$ and $\vec{F}$ , along the axis of rotation.

$\vec{\tau} = \vec{r} \times \vec{F}$

$|\tau| = r F \sin\theta$

Where $\vec{r}$ is the distance from the pivot to where the force is applied, and $\theta$ is the angle between the $\vec{r}$ and $\vec{F}$ vectors.

The Right-Hand Rule: To find the direction of the torque vector, point the fingers of your right hand in the direction of $\vec{r}$ (away from the pivot). Curl your fingers toward the direction of $\vec{F}$ . Your thumb will point in the direction of the Torque (either into the page or out of the page).

Concept First: Only the component of force that is perfectly perpendicular to the wrench handle ( $F_\perp = F\sin\theta$ ) actually creates torque. Pushing or pulling directly parallel to the wrench ( $\theta = 0^\circ$ or $180^\circ$ ) does exactly zero work to turn the bolt!

2. Newton’s Second Law for Rotation

Just as $\Sigma F = ma$ dictates linear dynamics, the sum of all torques dictates rotational dynamics. If an object is not in static equilibrium, a net torque will cause it to undergo angular acceleration ( $\alpha$ ).

$\Sigma \vec{\tau} = I \vec{\alpha}$

Solving Multi-Body Systems: When a hanging mass pulls a string wrapped around a massive pulley, the tension on both sides of the string is no longer equal. The difference in tension is what creates the net torque required to spin the heavy pulley!

3. Quick AP Practice

📚 Unit C5 Mastery Challenge

1. A force $\vec{F} = 3\hat{i} + 4\hat{j}$ Newtons is applied to an object at a position $\vec{r} = 2\hat{i} - 1\hat{j}$ meters from the origin. Calculate the torque vector $\vec{\tau}$ about the origin.

Check Answer

Set up the determinant for the cross product $\vec{r} \times \vec{F}$ :
$\tau = (r_x F_y - r_y F_x) \hat{k}$
$\tau = ((2)(4) - (-1)(3)) \hat{k}$
$\tau = (8 - (-3)) \hat{k} = \mathbf{11\hat{k} \text{ N}\cdot\text{m}}$ . (The torque points in the positive z-direction, out of the page).

2. A uniform rod of length $L$ and mass $M$ is attached to a wall by a frictionless hinge. It is held horizontally and released from rest. What is the initial angular acceleration ( $\alpha$ ) of the rod? (Note: $I_{hinge} = \frac{1}{3}ML^2$ )

Check Answer

Gravity acts at the center of mass, which is at a distance $r = L/2$ from the hinge.
Torque: $\tau = (L/2)(Mg)\sin(90^\circ) = \frac{MgL}{2}$ .

Use Newton’s 2nd Law for Rotation: $\tau = I\alpha$ .
$\frac{MgL}{2} = (\frac{1}{3}ML^2) \alpha$ .

Cancel $M$ and one $L$ :
$\frac{g}{2} = \frac{L}{3} \alpha \Rightarrow \mathbf{\alpha = \frac{3g}{2L}}$ .

1. Torque as a Cross Product

2. Newton’s Second Law for Rotation

3. Quick AP Practice

📚 Unit C5 Mastery Challenge

🔗 Next Steps