1. Physics
  2. Rotational Motion
  3. Torque

Rotational Motion · JEE & NEET Physics

Torque: notes and previous year questions

The turning effect of a force: τ = rF sin θ, its direction, τ = Iα, pulleys with mass, equilibrium, the torque of gravity, and sliding against tipping.

Torque in short

  • Torque is the turning effect of a force about an axis.
  • τ = rF sin θ: equally, r times the part of F across r, or the lever arm times F.
  • Torque is zero for a force along r, through the axis, or of zero size.
  • Torque is a vector, τ = r × F, along the axis by the right-hand rule.

1What is torque?

Push a door with the same force in different places. Near the hinge it barely moves; at the handle it swings open easily; pushing along the door toward the hinge, it does not turn at all. The turning effect of a force is called torque, τ\tau.

Torque depends on how big the force is, how far from the axis it acts, and in which direction it pushes. A bigger force does not always win: 10 N at 2 m gives 20 N m, but 100 N at 0.1 m gives only 10 N m.

2τ = rF sin θ

τ=rFsin⁡θ\tau = rF\sin\thetar: from the axis to where the force acts; θ: the angle between r and F. Unit: N m.
  • τ=rF⊥\tau = rF_\perp: only the part of the force across rr, Fsin⁡θF\sin\theta, turns; the part along rr just pulls on the axis.
  • τ=r⊥F\tau = r_\perp F: the lever arm r⊥=rsin⁡θr_\perp = r\sin\theta is the shortest distance from the axis to the line of the force.
  • Largest at θ=90°\theta = 90° (τ=rF\tau = rF); zero when θ=0°\theta = 0° or 180°180°, when r=0r = 0, or when F=0F = 0.

3The direction of torque

τ⃗=r⃗×F⃗\vec\tau = \vec r \times \vec FA cross product: perpendicular to both r and F, along the axis of turning.

Right-hand rule: curl the fingers of your right hand the way the body turns; your thumb points along the torque. Anticlockwise as you look at it: torque out of the page, toward you. Clockwise: into the page.

When adding torques in one plane, take anticlockwise as positive and clockwise as negative, and keep to that choice through the problem.

4τ = Iα

τnet=Iα\tau_{net} = I\alphaNewton's second law for rotation: F → τ, m → I, a → α.
W=τθ,P=τωW = \tau\theta, \qquad P = \tau\omegaWork and power of a torque, like W = Fs and P = Fv.

5A pulley with mass

Write F=maF = ma for each block and τ=Iα\tau = I\alpha for the pulley, with a=Rαa = R\alpha (the rope does not slip).

6Balancing torques

A rigid body stays at rest only if both hold:

∑F⃗=0and∑τ⃗=0\sum \vec F = 0 \quad\text{and}\quad \sum \vec\tau = 0The torques must balance about every axis, so choose the one that removes the most unknown forces.

A couple is two equal and opposite forces not on one line. The net force is zero, so the centre of mass stays put, but the torque is not: the body just spins. A rod on a smooth table pushed square to it at both ends, opposite ways: τ=FL\tau = FL and α=FLmL2/12=12FmL\alpha = \frac{FL}{mL^2/12} = \frac{12F}{mL}, while acm=0a_{cm} = 0.

7The torque of gravity

For torque, the whole weight of a body acts at its centre of mass: τg=Mg rcmsin⁡θ\tau_g = Mg\,r_{cm}\sin\theta.

A rod hinged at one end and held at 30° to the vertical by a horizontal force FF at its free end: torques about the hinge give FLcos⁡30°=MgL2sin⁡30°F L\cos30° = Mg\frac L2\sin30°, so F=Mg2tan⁡30°≈0.29MgF = \frac{Mg}{2}\tan30° \approx 0.29Mg.

8Slide or tip?

Push a block (side aa, mass mm) sideways at height hh:

  • It slides when F>μmgF > \mu mg.
  • It tips about its front edge when the push's torque beats the weight's: Fh>mga2Fh > mg\frac a2, so F>mga2hF > \frac{mga}{2h}.
  • Whichever limit is smaller happens first.

Summary

Key ideas

  • Torque is the turning effect of a force about an axis.
  • τ = rF sin θ: equally, r times the part of F across r, or the lever arm times F.
  • Torque is zero for a force along r, through the axis, or of zero size.
  • Torque is a vector, τ = r × F, along the axis by the right-hand rule.
  • Net torque gives angular acceleration: τ = Iα; torque does work τθ at power τω.
  • A pulley with mass needs different tensions on its two sides.
  • At rest: the forces and the torques (about any axis) must both balance.
  • A couple has no net force but a net torque: it spins a body in place.
  • The weight of a body acts at its centre of mass; a hinged rod's end starts falling at 3g/2.
  • A pushed block slides if F > μmg and tips if Fh > mg a/2, whichever comes first.

Every equation

Torque
τ=rFsin⁡θ\tau = rF\sin\theta
Lever arm form
τ=r⊥F=rF⊥\tau = r_\perp F = rF_\perp
Vector form
τ⃗=r⃗×F⃗\vec\tau = \vec r \times \vec F
Second law for rotation
τnet=Iα\tau_{net} = I\alpha
Work by a torque
W=τθW = \tau\theta
Power of a torque
P=τωP = \tau\omega
Rope on a pulley
a=Rαa = R\alpha
Two blocks, disc pulley
a=(m1−m2)gm1+m2+I/R2a = \frac{(m_1 - m_2)g}{m_1 + m_2 + I/R^2}
Block on a string round a disc
a=mgm+M/2a = \frac{mg}{m + M/2}
Equilibrium
∑F=0, ∑τ=0\sum F = 0,\ \sum\tau = 0
Couple on a rod
τ=FL\tau = FL
Torque of gravity
τg=Mg rcmsin⁡θ\tau_g = Mg\,r_{cm}\sin\theta
Hinged rod released
α=3g/2L\alpha = 3g/2L
Hinge force at release
R=Mg/4R = Mg/4
Sliding
F>μmgF > \mu mg
Tipping
F>mga2hF > \frac{mga}{2h}

Previous year questions with solutions

Real JEE and NEET questions on torque. Try each one before you open the solution.

Q1JEE Main 2026One correct option

A solid sphere of radius 4 cm and mass 5 kg is rotating (rotation axis is passing through the centre of the sphere) with an angular velocity of 1200 rpm . It is brought to rest in 10 s by applying a constant torque. The torque applied and the number of rotations it made before it comes to rest are ____\_\_\_\_ and ____\_\_\_\_ respectively.

  1. A0.128πNm,1000.128\pi \mathrm{Nm},100
  2. B0.0128πNm,500.0128\pi \mathrm{Nm},50
  3. C0.128πNm,500.128\pi \mathrm{Nm},50
  4. D0.0128πNm,1000.0128\pi \mathrm{Nm},100
Show answer and solution

Answer: Option D

First the sphere: I=25MR2=25×5×0.042=0.0032 kg m2I = \tfrac{2}{5}MR^{2} = \tfrac{2}{5} \times 5 \times 0.04^{2} = 0.0032\ \mathrm{kg\,m^{2}}, with the radius in metres. Its speed: 1200 rpm=1200×2π60=40π rad/s1200\ \mathrm{rpm} = \dfrac{1200 \times 2\pi}{60} = 40\pi\ \mathrm{rad/s}, all lost in 10 s10\ \mathrm{s}, so ∣α∣=4π rad/s2|\alpha| = 4\pi\ \mathrm{rad/s^{2}} and τ=I∣α∣=0.0032×4π=0.0128π N m\tau = I|\alpha| = 0.0032 \times 4\pi = 0.0128\pi\ \mathrm{N\,m}.

Then the turns: θ=12ω0t=12×40π×10=200π rad\theta = \tfrac{1}{2}\omega_{0}t = \tfrac{1}{2} \times 40\pi \times 10 = 200\pi\ \mathrm{rad}, which is 200π2π=100\dfrac{200\pi}{2\pi} = 100 rotations. Or count turns directly: it starts at 2020 turns a second and averages 1010, for 10 s10\ \mathrm{s}.

The trap is 5050: the average speed, 20π rad/s20\pi\ \mathrm{rad/s}, is already half of 40π40\pi, and halving it again loses a factor of two. 0.128π0.128\pi is ten times too big, a slip in squaring 0.040.04.

Q2NEET 2025One correct option

A uniform rod of mass 20 kg and length 5 m leans against a smooth vertical wall making an angle of 60∘{60}^{\circ } with it. The other end rests on a rough horizontal floor. The friction force that the floor exerts on the rod is (Take g=10m/s2g=10 m/s^{2})

  1. A200 N
  2. B2003N200\sqrt{3} N
  3. C100 N
  4. D1003N100\sqrt{3} N
Show answer and solution

Answer: Option D

The angle is given with the wall: 60∘60^{\circ} with the wall is 30∘30^{\circ} with the floor. The forces give N2=Mg=200 NN_{2} = Mg = 200\ \mathrm{N} and f=N1f = N_{1}; moments about the foot give N1Lsin⁡30∘=MgL2cos⁡30∘N_{1}L\sin 30^{\circ} = Mg\dfrac{L}{2}\cos 30^{\circ}. So f=N1=Mg2cot⁡30∘=1003 N≈173 Nf = N_{1} = \dfrac{Mg}{2}\cot 30^{\circ} = 100\sqrt{3}\ \mathrm{N} \approx 173\ \mathrm{N}. The length, 5 m5\ \mathrm{m}, cancels.

The trap is B, 2003 N200\sqrt{3}\ \mathrm{N}, from putting the weight at the top of the rod instead of its middle. Taking 60∘60^{\circ} as the angle with the floor gives 1003≈58 N\dfrac{100}{\sqrt{3}} \approx 58\ \mathrm{N}, which is not offered.

Q3JEE Advanced 2019One or more correct options

A thin and uniform rod of mass M and length L is held vertical on a floor with large friction. The rod is released from rest so that it falls by rotating about its contact-point with the floor without slipping. Which of the following statement(s) is/are correct, when the rod makes an angle 60∘{}^{\circ } with vertical? [g is the acceleration due to gravity]

  1. AThe angular acceleration of the rod will be 2gL\frac{2g}{L}.
  2. BThe normal reaction force from the floor on the rod will be Mg16\frac{Mg}{16}.
  3. CThe radial acceleration of the rod's center of mass will be 3g4\frac{3g}{4}.
  4. DThe angular speed of the rod will be 3g2L\sqrt{\frac{3g}{2L}}.
Show answer and solution

Answer: Options B, C, D

Energy about the foot: the centre drops L2(1−cos⁡60∘)=L4\dfrac{L}{2}(1 - \cos 60^{\circ}) = \dfrac{L}{4}, so MgL4=12⋅13ML2ω2Mg\dfrac{L}{4} = \tfrac{1}{2} \cdot \tfrac{1}{3}ML^{2}\omega^{2} and ω2=3g2L\omega^{2} = \dfrac{3g}{2L}: D is right.

Torque about the foot: α=MgL2sin⁡60∘13ML2=33g4L≈1.3gL\alpha = \dfrac{Mg\frac{L}{2}\sin 60^{\circ}}{\frac{1}{3}ML^{2}} = \dfrac{3\sqrt{3}g}{4L} \approx 1.3\dfrac{g}{L}, not 2gL\dfrac{2g}{L}: A is wrong.

The centre's acceleration along the rod, towards the foot: ω2L2=3g4\omega^{2}\dfrac{L}{2} = \dfrac{3g}{4}: C is right.

For the floor's push, take the downward parts of the centre's two accelerations. The part across the rod, αL2=33g8\alpha\dfrac{L}{2} = \dfrac{3\sqrt{3}g}{8}, points 30∘30^{\circ} from straight down, giving 33g8⋅32=9g16\dfrac{3\sqrt{3}g}{8} \cdot \dfrac{\sqrt{3}}{2} = \dfrac{9g}{16}. The part along the rod, 3g4\dfrac{3g}{4}, points 60∘60^{\circ} from straight down, giving 3g4⋅12=6g16\dfrac{3g}{4} \cdot \dfrac{1}{2} = \dfrac{6g}{16}. So the centre accelerates downwards at 15g16\dfrac{15g}{16}, and Mg−N=M15g16Mg - N = M\dfrac{15g}{16} gives N=Mg16N = \dfrac{Mg}{16}: B is right. At this moment the floor barely pushes at all.

The trap in B is to keep only one of the two parts of the centre's acceleration: the part across the rod alone gives N=7Mg16N = \dfrac{7Mg}{16}.

Practice questions, easy to hard

Three questions from the torque practice ladder: one easy, one medium, one hard.

Q4Numerical answer

A block of 1 kg1\ \mathrm{kg} hangs from a string wrapped round a uniform disc of mass 2 kg2\ \mathrm{kg} and radius 0.1 m0.1\ \mathrm{m}, which turns on a fixed, smooth axle. For a disc, I=12MR2I = \tfrac{1}{2}MR^{2}. The block is released from rest. Taking g=10 m/s2g = 10\ \mathrm{m/s^{2}}, find the tension in the string, in newtons.

Show answer and solution

Answer: 5 N

I=12×2×0.12=0.01 kg m2I = \tfrac{1}{2} \times 2 \times 0.1^{2} = 0.01\ \mathrm{kg\,m^{2}}, so ImR2=0.011×0.01=1\dfrac{I}{mR^{2}} = \dfrac{0.01}{1 \times 0.01} = 1 and a=101+1=5 m/s2a = \dfrac{10}{1 + 1} = 5\ \mathrm{m/s^{2}}. Then the block's own law gives the tension: T=m(g−a)=1×(10−5)=5 NT = m(g - a) = 1 \times (10 - 5) = 5\ \mathrm{N}.

Check with the pulley: TR=5×0.1=0.5 N mTR = 5 \times 0.1 = 0.5\ \mathrm{N\,m}, and Iα=0.01×50.1=0.5 N mI\alpha = 0.01 \times \dfrac{5}{0.1} = 0.5\ \mathrm{N\,m}. The trap is 10 N10\ \mathrm{N}, the block's weight: the string holds the block with less than its weight, because the block is accelerating downwards.

Q5One correct option

The floor can supply friction only up to μN2=μW\mu N_{2} = \mu W. The shallower the ladder, the more friction it needs; the least angle at which it can stand is the one where the friction it needs just equals μW\mu W. Any lower and it slips.

A uniform ladder rests against a smooth wall on a floor with μ=0.5\mu = 0.5. What is the least angle with the floor at which it can stand without slipping?

  1. A26.6∘26.6^{\circ}
  2. B45∘45^{\circ}
  3. C60∘60^{\circ}
  4. D63.4∘63.4^{\circ}
Show answer and solution

Answer: Option B

At the least angle, W2cot⁡θ=μW\dfrac{W}{2}\cot\theta = \mu W, so cot⁡θ=2μ=1\cot\theta = 2\mu = 1 and θ=45∘\theta = 45^{\circ}. Notice that WW cancels: a heavy ladder and a light one slip at the same angle. The traps: D, 63.4∘63.4^{\circ}, has tan⁡θ=2\tan\theta = 2 — it drops the half, putting the weight at the top. A, 26.6∘26.6^{\circ}, has tan⁡θ=0.5\tan\theta = 0.5, setting tan⁡θ\tan\theta equal to μ\mu as for a block on an incline. Check at 45∘45^{\circ}: the friction needed is W2cot⁡45∘=0.5W\dfrac{W}{2}\cot 45^{\circ} = 0.5W, exactly the most the floor can give.