# How to Calculate the Torque of a Rotating Roll

Written by gary damico
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Torque is a concept frequently used in mechanics. It is associated with objects that spin around some fixed axis -- whether a marble rolling down a hill or the Moon around the Earth. In order to calculate torque, you need to find the product of the moment of inertia of the object about this axis and the rate of change of angular velocity, also called angular acceleration. The moment of inertia depends not only on the location of the axis, but also on the shape of the object. For a "rotating roll," we will assume that the roll is a perfect cylinder and that its centre of mass is at its geometric centre. Also, we will neglect air resistance -- as with many problems in physics, these assumptions neglect some real-world complications but are necessary to create a soluble problem!

Skill level:
Moderate

### Things you need

• Mass of the roll
• Rate of change of angular velocity
• Timepiece

## Instructions

1. 1

Review your initial definitions. The Moment of Inertia is given by I = I(0) + mx^2, where I(0) is the moment of inertia around an axis through the centre of the object and x is the distance from the axis of rotation to the centre of mass. Note that if the axis we are studying is through the centre of mass, then the second term in the equation above disappears.

For a cylinder, I(0) = mr^2/2 where r is the radius of the cylinder and m its mass. So for example, if the axis of rotation is through the centre of mass, then I = I(0) = mr^2/2 and if the axis of rotation is halfway to the edge, then I = I(0) + mx^2 = mr^2/2 + m(r/2)^2 = 3mr^2/4.

2. 2

Find the angular velocity. Angular velocity ω (lower case omega) is a measure of the speed of rotation and is measured in radians per second. You can calculate this (i) directly by determining the number of rotations the cylinder makes in a given time; or (ii) by finding the velocity v (distance per time) of any point on the cylinder and dividing by the distance from that point to the centre of mass; in the latter approach, ω = v/r.

3. 3

Find the angular acceleration. Torque depends on angular acceleration α (lower case alpha) which is the rate of change of angular velocity ω, so we must find the CHANGE in ω for the time period we are considering. Then α = "ω/"t. For example if the roll goes from ω = 6 radians/sec to ω = 0 radians/sec in three seconds, then α = "ω/"t = 6/3 = 2 radians/sec^2.

4. 4

Calculate torque. Torque τ = Iα. For example if our cylinder has a mass of 20 grams = 0.02kg and a radius of 5cm = 0.05m, and it is rotating about an axis through its centre, then I = mr^2 = (.02)(.05)² = 0.00005 = 5e-5 kgm^2. And if we use the α from Step 3, then torque τ = Iα = 5e-5 (2) = 1e-4 meter-newtons.

#### Tips and warnings

• The moment of inertia is at a minimum when the axis of rotation is through the centre of the roll, and at a maximum when the axis is tangent to its circumference. Therefore, I will always lie between I(0) = mr^2/2 and mr^2/2 + mr^2 = 3mr^2/2.
• The easiest mistake to make in this calculation is to use the wrong units of measure. When using the metric system, distances must be in meters, time in seconds, mass in kilograms, and ω in radians per second (remember that 2π radians = 360 degrees). Then the torque will be measured in meter-newtons.
• For the English system, distances must be measured in feet, time in seconds, mass in slugs (NOT pounds (1 slug = 14.6kg)), and ω in radians per second. Then the torque will be measured in foot-pounds.
• Also, remember that torque depends on the CHANGE in ω, not ω itself. So in particular, a roll with constant ω will have zero torque.
• Third, remember that we must make certain approximations as described above. The effect of these approximations on the final result is negligible for most situations, but can be significant if the roll is not of nearly uniform density or experiences strong friction along its axis of rotation.

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