Torque and Equations of Motion | JEE Notes

A force can accelerate the centre of mass, change the rotation, or do both. Its rotational effect is determined by where its line of action lies relative to the chosen axis—not merely by its magnitude.

Choose a point $O$ , draw the position vector $𝐫$ from $O$ to the point of application, and define

𝛕_{O} = 𝐫 \times 𝐅, | τ_{O} | = r F \sin ϕ = F d_{⊥} .

Here $d_{⊥}$ is the shortest distance from the axis to the line of action. Extend that line if necessary. A force applied far from the hinge can still have zero torque if it points directly towards the hinge.

Watch the animation: Torque and the line of action

For motion in the $x y$ -plane, take counterclockwise as positive, corresponding to $+ z$ out of the page. Then $τ_{z} = x F_{y} - y F_{x}$ . The right-hand rule gives the same sign. Do not attach a permanent “positive” sign to gravity or friction: either force can produce either sense of rotation, depending on the geometry.

Explore in 3D: Torque as a cross product

A couple consists of two equal, opposite, parallel forces with distinct lines of action. Its net force is zero; its torque magnitude is $F d$ , where $d$ separates the lines. Because shifting the origin changes torque by a term proportional to the resultant force, a couple has the same torque about every origin. A couple can accelerate rotation without accelerating the centre of mass.

Which rotational equation is actually valid?

For a rigid body rotating about a stationary axis with constant axial moment of inertia, take the component along that axis:

\sum τ_{ext, z} = I_{z} α .

For general planar rigid-body motion, use the pair

\sum 𝐅_{e x t} = M 𝐚_{C}, \sum τ_{C, z} = I_{C} α .

The centre-of-mass equation for torque remains valid even when the centre of mass accelerates. By contrast, the unrestricted three-dimensional vector statement $𝛕 = I 𝛂$ is generally false: inertia is a tensor, and angular momentum need not be parallel to angular velocity. The scalar equations here apply to the stated planar or fixed-axis motion.

Beyond JEE: moments about a moving point

Can you take moments about a convenient moving point $A$ ? Yes, but with care. If angular momentum is defined using velocities measured in an inertial frame, then

\frac{d 𝐋_{A}}{d t} = 𝛕_{A} - 𝐯_{A} \times 𝐏, 𝐏 = M 𝐯_{C} .

The extra term vanishes for a fixed inertial point, or for $A = C$ . It does not vanish automatically for a moving contact point. Also, even when it vanishes, $𝐋_{A}$ need not equal $I_{A} 𝛚$ . Choose the centre of mass or a genuinely fixed pivot until you can justify a different origin.

Checkpoint: a free rod is pushed at its centre. Will it rotate?

If it starts with no rotation and the only applied in-plane force passes through its centre, the torque about the centre is zero, so its angular acceleration is zero. The rod translates. Taking torque about some distant fixed point may give a nonzero result, but that torque changes the orbital angular momentum of the translating centre. It is not evidence that the rod must spin. This is why the origin must appear explicitly in your equations.

Beyond JEE: sliding a force along its line

A related shortcut has a strict limit. Sliding a force along its own line of action preserves its resultant force and moment, so it preserves the rigid body's external acceleration. Moving that force to a parallel but displaced line changes the torque. To make that relocation equivalent, you must add a compensating couple. The body's internal stress distribution can change even when its external force and moment are unchanged; rigid-body mechanics here does not calculate those stresses.

A force that turns a flywheel clockwise

A flywheel rotates about a fixed frictionless $z$ -axis through $O$ . At one instant, a force $𝐅 = (6 \hat{𝐢} - 8 \hat{𝐣}) N$ acts at $𝐫 = (0.30 \hat{𝐢} + 0.40 \hat{𝐣}) m$ . Its axial inertia is $0.60 k g m^{2}$ . Find the signed torque and angular acceleration. No other force supplies an axial torque.

Hint: calculate the cross product before interpreting the motion

Use $x F_{y} - y F_{x}$ . A negative angular acceleration does not, by itself, mean that the wheel is slowing down.

Full solution

τ_{z} = (0.30) (- 8) - (0.40) (6) = - 4.80 N m, α = \frac{- 4.80}{0.60} = - 8.00 r a d s^{- 2} .

The torque is clockwise. Its perpendicular lever arm is $d_{⊥} = | τ | / F = 4.8 / 10 = 0.48 m$ . If the wheel currently rotates counterclockwise, it slows; if it rotates clockwise, its speed increases. The axle force acts through $O$ and contributes no torque about this axis.

Check: $d_{⊥} = 0.48 m$ is smaller than $| 𝐫 | = 0.50 m$ , as required. Torque determines the change in angular velocity, not its current value.

Why a released hinged rod does not fall freely

A uniform rod of mass $3 k g$ and length $2 m$ is hinged at its left end and released from rest while horizontal. Find its initial angular acceleration and hinge reaction. Use $g = 10 m s^{- 2}$ .

Hint: take torque about the hinge, then use the centre of mass

The hinge reaction has zero hinge torque but is essential in $\sum 𝐅 = M 𝐚_{C}$ .

Full solution

Take counterclockwise as positive. The fixed hinge gives

I_{O} = \frac{M L^{2}}{3} = 4 k g m^{2}, τ_{O} = - M g \frac{L}{2} = - 30 N m, α = - 7.5 r a d s^{- 2} .

Initially $ω = 0$ , so there is no centripetal acceleration. The centre has downward tangential acceleration $| α | L / 2 = 7.5 m s^{- 2}$ . With upward positive,

R_{y} - M g = - M (7.5), R_{y} = 7.5 N, R_{x} = 0 .

The result is a purely upward initial hinge reaction. A hinge is not necessarily supporting the full weight when the body accelerates.

Check: Generally $| α | = 3 g / (2 L)$ , $| a_{C} | = 3 g / 4$ , and $R_{y} = M g / 4$ . The rod's centre falls more slowly than a freely falling particle at that instant.

Exam trap: A zero torque about one point does not imply zero resultant force. Conversely, a zero resultant force does not imply zero torque. Equilibrium requires both.

Vector form, as JEE Main asks it

Most torque questions in JEE Main are one determinant. Write $\vec{r}$ on the top line and $\vec{F}$ below it, because swapping them flips the sign:

\vec{τ} = | \begin{matrix} \hat{i} & \hat{j} & \hat{k} \\ x & y & z \\ F_{x} & F_{y} & F_{z} \end{matrix} |

For a force acting at point $A$ , about point $B$ , use $\vec{r} = {\vec{r}}_{A} - {\vec{r}}_{B}$ . The same determinant with $\vec{p}$ gives $\vec{L}$ . A particle moving in a straight line has constant $\vec{L}$ about any point if no force acts: $L = m v d_{⊥}$ .

A spinning disc on a rough table

Friction acts on every ring at its own radius, so the torque is not $μ M g R$ . Summing rings:

τ = \int_{0}^{R} μ g \frac{2 M}{R^{2}} r^{2} d r = \frac{23}{μ} M g R .

The stopping time from

ω

t = \frac{3 ω R}{4 μ g}

Torque: choose the axis before writing the equation

In 60 seconds

Which rotational equation is actually valid?

A force that turns a flywheel clockwise

Why a released hinged rod does not fall freely

Vector form, as JEE Main asks it

A spinning disc on a rough table