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Torque drives rotation

Torque is the rotational version of force, and tau = I alpha is Newton's second law spun on its axis.

Torque

Torque is force made rotational: tau = r F sin(phi), where r runs from the axis to where the force acts and phi is the angle between r and F. Only the component of force perpendicular to the lever arm turns anything — pushing straight along a wrench handle, through the bolt, produces zero torque. The lever arm r sin(phi) is the perpendicular distance from the axis to the force's line of action.

\tau = rF\sin\phi

Second law, rotated

Net torque equals moment of inertia times angular acceleration — the exact twin of F = ma. A given torque spins up a flywheel slowly if the flywheel is heavy or its mass sits far from the axis. For problems with a rope over a pulley, apply F = ma to the hanging mass, tau = I alpha to the pulley, and let the no-slip rope connect them: a = r alpha.

\tau_{\text{net}} = I\alpha

Worked example

A frictionless pulley is a uniform disk, mass 2.0 kg and radius 0.10 m, with a 1.0 kg mass hanging from the wrapped rope. For the mass: mg - T = ma. For the pulley: I = (1/2)MR^2 = 0.010 kg m^2, and with alpha = a/r the pulley equation gives T = I a/r^2 = 1.0a. Substituting into mg - T = ma: 9.8 - 1.0a = 1.0a, so a = 4.9 m/s^2 and T = 4.9 N. The mass falls at half of g because half the energy budget goes into spinning the disk.

a = \frac{mg}{m + M/2} = \frac{9.8}{1.0 + 1.0} = 4.9\ \text{m/s}^2

Flashcard

\tau = rF\sin\phi

Torque from an off-axis push

Only the force component perpendicular to the lever arm causes rotation, which is where the sine comes from. Maximum torque means pushing at a right angle to r, and pushing along r gives zero.

Check your understanding

A 24 N force is applied at the rim of a wheel of radius 0.25 m, at an angle of 30 degrees to the spoke (the radius direction). What is the torque about the axle?

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