Torque Practice Problems- Master Rotational Dynamics

What Is Torque, Really?

Torque is the rotational equivalent of force. While force makes things accelerate in a straight line, torque makes things spin. It's that simple.

The formula is straightforward:

τ = r × F × sin(θ)

Where:

• τ is torque (measured in Newton-meters, or N·m)
• r is the lever arm distance from the pivot point to where force is applied
• F is the force applied
• θ is the angle between the force vector and the lever arm

When the force is perpendicular to the lever arm, sin(90°) = 1, and the formula simplifies to τ = rF. This is the most common scenario you'll encounter in practice problems.

The Key Relationship: Torque and Angular Acceleration

Newton's Second Law for rotation states:

τ_net = Iα

This is your bread and butter. The net torque on an object equals its moment of inertia times its angular acceleration. Every rotational dynamics problem ultimately comes back to this equation.

Moment of Inertia for Common Shapes

• Point mass: I = mr²
• Solid cylinder/disk: I = ½mr²
• Hollow cylinder: I = mr²
• Solid sphere: I = ⅖mr²
• Thin rod (about center): I = ⅟₁₂mr²

Commit these to memory. You'll use them constantly.

Practice Problem 1: The Wrench

A 30 cm wrench is used to tighten a bolt. If 200 N of force is applied perpendicular to the wrench handle, what torque is produced?

Solution

Given: r = 0.30 m, F = 200 N, θ = 90°

Since the force is perpendicular:

τ = rF = (0.30 m)(200 N) = 60 N·m

That's it. Plug and chug. The answer is 60 Newton-meters of torque.

Practice Problem 2: Two Forces on a Disk

A solid disk (mass = 4 kg, radius = 0.5 m) rotates about its center. Two forces are applied at the rim:

• Force A: 10 N tangent to the disk
• Force B: 15 N tangent to the disk (opposite direction)

What is the angular acceleration of the disk?

Solution

First, calculate the moment of inertia for a solid disk:

I = ½mr² = ½(4 kg)(0.5 m)² = 0.5 kg·m²

Net torque:

τ_net = (10 N)(0.5 m) - (15 N)(0.5 m) = 5 - 7.5 = -2.5 N·m

The negative sign indicates the disk accelerates in the direction of the 15 N force.

Angular acceleration:

α = τ_net / I = 2.5 / 0.5 = 5 rad/s² (in the direction of the larger force)

Practice Problem 3: The Rolling Object

A solid sphere (mass = 3 kg, radius = 0.2 m) rolls down a 30° incline without slipping. Calculate the linear acceleration down the ramp.

Solution

For rolling without slipping, use the relationship: a = αr

The moment of inertia for a solid sphere: I = ⅖mr²

Torque about the contact point causes rotation. The component of gravity parallel to the incline creates torque:

τ = (mg sin θ)r

Using the parallel axis theorem or analyzing forces directly:

a = (⅗) g sin θ = (⅗)(9.8)(0.5) = 2.94 m/s²

Compare this to a sliding block: a = g sin θ = 4.9 m/s². Rolling objects accelerate slower because energy goes into rotation, not just translation.

Practice Problem 4: Atwood's Machine with a Pulley

A massive pulley (I = 0.5 kg·m², radius = 0.2 m) has masses m₁ = 2 kg and m₂ = 4 kg hanging on either side. Find the angular acceleration of the pulley.

Solution

Free body diagram time. For m₁ (going up): T₁ - m₁g = m₁a

For m₂ (going down): m₂g - T₂ = m₂a

For the pulley: (T₂ - T₁)r = Iα, and α = a/r

Substitute α = a/r into the pulley equation:

(T₂ - T₁)r = I(a/r)

T₂ - T₁ = Ia/r²

From the first two equations:

T₁ = m₁g + m₁a

T₂ = m₂g - m₂a

T₂ - T₁ = m₂g - m₂a - m₁g - m₁a = (m₂ - m₁)g - (m₂ + m₁)a

Set equal to Ia/r²:

(m₂ - m₁)g - (m₂ + m₁)a = Ia/r²

Solve for a:

a[(m₂ + m₁) + I/r²] = (m₂ - m₁)g

a = (m₂ - m₁)g / [(m₂ + m₁) + I/r²]

a = (4-2)(9.8) / [(4+2) + 0.5/0.04]

a = 19.6 / [6 + 12.5] = 19.6 / 18.5 = 1.06 m/s²

Then α = a/r = 1.06 / 0.2 = 5.3 rad/s²

Common Mistakes That Cost You Points

• Using the wrong r: Always use the perpendicular distance from the pivot to where force is applied. Not the full length if force isn't perpendicular.
• Forgetting the mass of rotating objects: In problems with pulleys or wheels, don't ignore their rotational inertia.
• Sign errors: Torque can be positive or negative depending on direction. Be consistent with your sign convention.
• Mixing units: Convert everything to meters, kilograms, seconds before plugging in.
• Assuming no friction: Problems often specify "massless, frictionless" pulleys. When they don't, friction matters.

Comparing Rotational Dynamics Formulas

How to Approach Any Torque Problem

1. Identify the pivot point. This is where you'll calculate torques from. Sometimes you can choose any point—pick one that simplifies the math.
2. Draw a free body diagram. Show all forces, all distances from the pivot.
3. Calculate each torque. Use τ = rF sin θ. Pay attention to signs.
4. Apply τ_net = Iα. This is your master equation.
5. Solve for the unknown. Isolate the variable and calculate.
6. Check units. Torque should be N·m, angular acceleration rad/s², etc.

When to Use Energy vs. Torque

Torque and angular acceleration problems often have energy counterparts. If the problem asks for speed at a certain position, or maximum height reached, consider using conservation of energy:

Initial energy = Final energy

Rotational kinetic energy (½Iω²) + translational kinetic energy (½mv²) + potential energy (mgh) should equal the initial total.

Use torque when you need acceleration, angular acceleration, or angular velocity as functions of time. Use energy when you need speeds or positions without time dependence.