Course 6.3Lesson 3 of 3

Rolling motion

Rolling without slipping is moving along at $v$ plus turning at $\omega = \dfrac{v}{R}$

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Try the code and the simulations beside the lesson.

Moving and turning at once

A rolling wheel does two things together: its centre moves along at $v$ , and it turns at $\omega$ . Rolling without slipping means the point touching the ground does not skid, so

\displaystyle v = \omega R

A wheel that rolls to the right on a horizontal road without slipping. The centre moves with the velocity v, shown as a blue arrow. The top point moves with the velocity 2 v, shown as a longer red arrow. The point where the wheel touches the road is marked in green, and it is at rest for that instant.

At the bottom the two motions cancel: that point is at rest for an instant. At the top they add: it moves at $2v$ .

Energy of rolling

\displaystyle K = \tfrac12 Mv^2 + \tfrac12 I\omega^2 = \tfrac12 Mv^2\left(1 + \frac{I}{MR^2}\right)

The number $\dfrac{I}{MR^2}$ is $\tfrac25$ for a solid ball, $\tfrac12$ for a disc or solid cylinder, $\tfrac23$ for a hollow ball and $1$ for a ring.

Rolling down a slope

Friction at the contact point makes the body turn, but since that point does not slide, friction does no work. Energy gives the speed after a drop $h$ :

\displaystyle v = \sqrt{\frac{2gh}{1 + I/MR^2}}, \qquad a = \frac{g\sin\theta}{1 + I/MR^2}

A cylinder on a rough slope that makes an angle theta with the horizontal. The weight m g points straight down. The normal reaction N is at a right angle to the slope. The friction f points up the slope at the point of contact. A dashed blue arrow shows the acceleration a of the centre, down the slope.

Race them.

You should find the same order at every angle: ball, then disc, then ring, and all of them behind a frictionless sliding block.

Watch out. Mass and radius cancel. A huge ring and a tiny ring arrive together; only the shape, through $\dfrac{I}{MR^2}$ , matters.

A worked example

A hollow ball ( $I = \tfrac23 MR^2$ ) rolls from rest down a slope, dropping $1.2\ \text{m}$ .

Energy. $Mgh = \tfrac12 Mv^2\left(1 + \tfrac23\right) = \tfrac56 Mv^2$ .
Speed. $v = \sqrt{\dfrac{6gh}{5}} = \sqrt{\dfrac{6 \times 9.8\ \text{m/s}^2 \times 1.2\ \text{m}}{5}} = 3.76\ \text{m/s}$ .
Compare. A sliding block would reach $\sqrt{2gh} = 4.85\ \text{m/s}$ .

Think first. Write $K$ for the shape you have, using $\dfrac{I}{MR^2}$ , then use energy for speeds and $a = \dfrac{g\sin\theta}{1 + I/MR^2}$ for times.

Practise with new numbers

Try it in Python

The program races four shapes down a $3.0\ \text{m}$ slope.

Remember these

Rolling without slipping: $v = \omega R$ ; bottom point at rest, top at $2v$ .
$K = \tfrac12 Mv^2\left(1 + \dfrac{I}{MR^2}\right)$ .
Down a slope: $a = \dfrac{g\sin\theta}{1 + I/MR^2}$ ; ball beats disc beats ring.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

A car moves at $20\ \text{m/s}$ . How fast is the bottom of a tyre moving over the road? The top?

What fraction of a rolling disc's kinetic energy is in its turning?

A solid ball and a hollow ball of the same mass and size roll down a slope. Which wins?

A disc rolls down a slope that drops $0.90\ \text{m}$ . How fast is it moving at the bottom? ( $g = 9.8\ \text{m/s}^2$ .)

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Problems (4)

A solid cylinder of mass 5.0 kg and radius 0.12 m rolls without slipping on a level floor, with its…Easy · High School (11–12) · by StemPy Lab

A solid sphere is let go from rest at the top of a slope at 25°, 4.0 m long, and rolls without…Medium · High School (11–12) · by StemPy Lab

A ring, a disc and a solid sphere, all of different masses and sizes, roll from rest without…Medium · High School (11–12) · by StemPy Lab

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