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Turning faster: τ = Iα

Forces and motion · level 8, lesson 3 of 6

A question to start

A rover's wheel motor gives a torque through a gearbox. How quickly does the wheel spin up, and how hard can it push the rover forward before the wheel slips on Mars?

The idea

The rotational form of Newton's second law is

\[\tau_\text{net} = I\alpha,\]

where \(\alpha\) is the angular acceleration, in rad/s², the rate of change of \(\omega\). The table shows how closely rotation copies straight-line motion:

Straight line Rotation
Position \(x\) Angle \(\theta\)
Velocity \(v\) Angular velocity \(\omega\)
Acceleration \(a\) Angular acceleration \(\alpha\)
Mass \(m\) Moment of inertia \(I\)
Force \(F\) Torque \(\tau\)
\(F = ma\) \(\tau = I\alpha\)
\(\tfrac12 mv^2\) \(\tfrac12 I\omega^2\)

A driven wheel on the ground: the motor applies a torque \(\tau\) to the wheel. Friction from the ground acts at the rim, pushing the wheel (and so the rover) forward with \(F\), and twisting the wheel back with \(FR\). For the wheel, \(\tau - FR = I\alpha\); for the rover, \(F = Ma\). If the wheel's own inertia is small, \(F \approx \tau/R\).

τ from the motorF, frictionR

How big can \(F\) be? No bigger than static friction, \(\mu_s N\). On Mars, the wheel presses down with only 0.38 of its Earth load, so for the same motor, wheels slip at 0.38 of the torque they could use on Earth. A rover motor geared for Earth would spin its wheels on Mars; the torque must be limited to what the ground can take.

Worked examples

1. Spinning up a flywheel. \(I = 2.25\) kg m², torque 9 N m. \(\alpha = 9/2.25 = 4\) rad/s². From rest to 300 rad/s: \(300/4 = 75\) s.

2. The torque limit. A 1,000 kg rover has six wheels of radius 0.26 m, each carrying a sixth of the weight. Each wheel's load on Mars: \(1000 \times 3.71 / 6 = 618\) N. With \(\mu_s = 0.5\), the most grip per wheel is 309 N, and the torque at which a wheel starts to slip is \(309 \times 0.26 = 80\) N m. On Earth it would be 2.6 times more.

3. A rolling race, again. For a disc rolling down a slope, \(\tau = FR\) about its centre and \(I = \tfrac12 MR^2\) give \(F = \tfrac12 Ma\); with \(Mg\sin\theta - F = Ma\), \(a = \tfrac23 g\sin\theta\), as in the last lesson.

Try it

  1. A torque of 12 N m acts on a wheel with \(I = 0.6\) kg m². Find \(\alpha\).
  2. How long does it take to reach 50 rad/s from rest?
  3. A wheel of radius 0.5 m carries 400 N on Mars with \(\mu_s = 0.6\). What is the most torque it can use without slipping?
  4. Why does the same motor slip on Mars when it would not on Earth?
Answers
  1. \(12/0.6 = 20\) rad/s².
  2. \(50/20 = 2.5\) s.
  3. \(0.6 \times 400 \times 0.5 = 120\) N m.
  4. The wheel presses down less, so the ground's grip, \(\mu_s N\), is smaller.

Common mistake

Mixing revolutions and radians. \(\tau = I\alpha\) needs \(\alpha\) in rad/s². One revolution is \(2\pi\) rad, and 1 rpm is \(2\pi/60 = 0.105\) rad/s.

Where it leads