Rotational Dynamics: τ = Iα

A pulley that fights back

In every pulley problem you have solved so far, the cord tension was the same on both sides. That was never a law. It was an assumption (that the pulley had no mass) and a massless pulley needs no net torque to spin up, so the two pulls on it must balance.

Real pulleys have mass. Hang a brick from a cord wrapped around a heavy flywheel and the brick falls slowly, with the cord pulling on it well below its full weight. Something is absorbing part of the fall. To say how much, you need a law for how hard it is to change a spinning object's rate of spin. From moment of inertia you know $I$ measures how much a body resists being spun; from linear dynamics you know $F = ma$. The two fit together in a way you can derive rather than memorise.

Torque: the part of a force that twists

Push a door on the handle and it swings. Push with the same force right next to the hinge and almost nothing happens. Push straight at the hinge and nothing happens at all. So the turning effect of a force depends on the force, on where you apply it, and on the direction you push.

Let $\vec r$ be the vector from the pivot to the point where the force acts, with length $r$ in metres, and let $\theta$ be the angle between $\vec r$ and the force $\vec F$ in newtons. Split $\vec F$ into a piece along $\vec r$ and a piece perpendicular to it. The piece along $\vec r$ pulls the body toward or away from the pivot; the pivot absorbs it and nothing turns. Only the perpendicular piece, of size $F\sin\theta$, twists, and it should count for more when applied further out. So the turning effect is

$\tau = r F \sin\theta$

and it is called torque, measured in newton-metres. Read it two ways. Group it as $r\,(F\sin\theta)$: the perpendicular part of the force times the distance. Or group it as $(r\sin\theta)F$: the whole force times the lever arm, the perpendicular distance from the pivot to the line the force acts along. Same number; use whichever is easier to see. Torque is zero at $\theta = 0$ or $180^\circ$ and largest at $\theta = 90^\circ$, where $\tau = rF$. Pick one sense of rotation as positive and give each torque a sign.

Why torque equals I times alpha

Start with the simplest spinning thing there is: a single point mass $m$ stuck to the end of a massless rod of length $r$, free to swing about a pivot. Push it perpendicular to the rod with a force $F$. Newton's second law still holds for that mass, in the direction it is free to move:

$F = m a_t$

where $a_t$ is its acceleration along the circle, called the tangential acceleration. But the mass is going round a circle of fixed radius, so its tangential acceleration and the angular acceleration $\alpha$ (in rad/s²) are locked together by $a_t = r\alpha$. Substitute:

$F = m r \alpha$

Now multiply both sides by $r$. On the left you get $rF$, which is the torque, since the push was perpendicular. On the right you get $mr^2\alpha$, and $mr^2$ is exactly the moment of inertia of a point mass at radius $r$. So

$\tau = I\alpha$

For an extended body, do this for every particle and add. Internal forces come in equal and opposite pairs along the line joining the particles, so their torques cancel and only external torques survive; every particle shares the same $\alpha$ because the body is rigid. You get the same equation with $I$ now the moment of inertia of the whole body about the axis you chose:

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

Read it the way you read $F = ma$. Double the torque and you double the angular acceleration. Double $I$, by moving the same mass further out, and you halve it for the same torque. Zero net torque means constant $\omega$, not zero $\omega$.

Two more results fall out. If a torque turns a body through a small angle $d\theta$, the force's application point moves $ds = r\,d\theta$, so $dW = F_\perp\,ds = (rF_\perp)\,d\theta$, giving $W = \tau\theta$ for a constant torque and $P = \tau\omega$ for the power: the mirrors of $W = Fd$ and $P = Fv$. A torque of 8.0 N·m through 25 rad does 200 J; 14 N·m at 50 rad/s delivers 700 W.

Worked example 1: spinning up a grindstone

A grindstone with $I = 0.60\ \mathrm{kg\,m^2}$ starts from rest and is driven by a constant torque of $3.0\ \mathrm{N\,m}$. How fast is it turning after $8.0\ \mathrm{s}$?

Find the angular acceleration first:

$\alpha = \frac{\tau}{I} = \frac{3.0\ \mathrm{N\,m}}{0.60\ \mathrm{kg\,m^2}} = 5.0\ \mathrm{rad/s^2}$

The torque is constant, so $\alpha$ is constant, and the rotational version of $v = v_0 + at$ applies:

$\omega = 0 + (5.0)(8.0) = 40\ \mathrm{rad/s}$

Those two lines answer a family of problems. A flywheel with $I = 5.0\ \mathrm{kg\,m^2}$ spinning at $40\ \mathrm{rad/s}$ stopped in $10\ \mathrm{s}$ needs $\alpha = -40/10 = -4.0\ \mathrm{rad/s^2}$, so $\tau = (5.0)(4.0) = 20\ \mathrm{N\,m}$ opposing the spin.

Brakes need one extra step to get the torque. A pad pressed against a disk of radius $0.30\ \mathrm{m}$ with a normal force of $200\ \mathrm{N}$ and $\mu_k = 0.35$ produces a friction force $f = \mu_k N = (0.35)(200) = 70\ \mathrm{N}$, tangent to the rim, so $\tau = (0.30)(70) = 21\ \mathrm{N\,m}$. With $I = 1.2\ \mathrm{kg\,m^2}$ that is $\alpha = 21/1.2 = 17.5\ \mathrm{rad/s^2}$, and stopping from $50\ \mathrm{rad/s}$ takes $t = 50/17.5 = 2.86\ \mathrm{s}$.

Worked example 2: the pulley with mass

Now the brick and the flywheel. A $3.0\ \mathrm{kg}$ block hangs from a cord wrapped over a pulley that is a solid disk of mass $M = 2.0\ \mathrm{kg}$ and radius $R = 0.20\ \mathrm{m}$. Find the block's acceleration and the tension.

On blank paper, write down three things before touching any numbers: Newton's second law for the block, $\tau = I\alpha$ for the pulley, and the constraint that ties them together.

The block feels gravity down and tension up. Taking down as positive,

$mg - T = ma$

The pulley feels the cord pulling tangentially at its rim with the same $T$ (the cord does not stretch or slip), so the torque on it is $TR$, and

$TR = I\alpha, \qquad I = \tfrac12 MR^2$

The constraint: the cord does not slip on the rim, so the speed of the cord equals the speed of the rim, $v = \omega R$, and differentiating, $a = \alpha R$, so $\alpha = a/R$. Substitute that into the pulley equation:

$TR = \tfrac12 MR^2\cdot\frac{a}{R} = \tfrac12 MRa \quad\Longrightarrow\quad T = \tfrac12 Ma$

Tension is no longer whatever it needs to be; it is set by how hard you are spinning up the pulley. Put it into the block equation:

$mg - \tfrac12 Ma = ma \quad\Longrightarrow\quad a = \frac{mg}{m + \tfrac12 M}$

With numbers, $a = (3.0)(9.8)/(3.0 + 1.0) = 29.4/4.0 = 7.35\ \mathrm{m/s^2}$. The tension follows: $T = \tfrac12 Ma = (1.0)(7.35) = 7.35\ \mathrm{N}$, well under the block's weight of 29.4 N.

Notice the shape of the answer. The pulley contributes $I/R^2 = \tfrac12 M$ to the denominator, as if the system carried an extra $\tfrac12 M$ of mass that has to be accelerated. That "effective mass" $I/R^2$ is worth carrying in your head.

When the cord runs over a table

Add a second block and only the bookkeeping changes. A $4.0\ \mathrm{kg}$ block sits on a table with $\mu_k = 0.20$, pulled by a cord over a solid-disk pulley of mass $1.0\ \mathrm{kg}$ to a $3.0\ \mathrm{kg}$ hanging block. Find the acceleration.

The trap: the tensions on the two sides of the pulley are not equal. If they were, the net torque on the pulley would be zero and it would never spin up. Call them $T_1$ on the table side and $T_2$ on the hanging side.

Table block, horizontal, with normal force $N = m_1g$ so friction is $f = \mu_k m_1 g = (0.20)(4.0)(9.8) = 7.84\ \mathrm{N}$:

$T_1 - f = m_1 a$

Hanging block:

$m_2 g - T_2 = m_2 a$

Pulley, with the two cords pulling at the rim in opposite senses:

$T_2 R - T_1 R = I\alpha = \tfrac12 MR^2\cdot\frac{a}{R} \quad\Longrightarrow\quad T_2 - T_1 = \tfrac12 Ma$

Three equations, three unknowns. Add them all together and both tensions cancel:

$m_2 g - f = \left(m_1 + m_2 + \tfrac12 M\right)a$

$a = \frac{(3.0)(9.8) - 7.84}{4.0 + 3.0 + 0.5} = \frac{29.4 - 7.84}{7.5} = \frac{21.56}{7.5} = 2.87\ \mathrm{m/s^2}$

The pattern generalises: driving forces on top, total inertia including $I/R^2$ on the bottom. An Atwood machine with $4.0$ and $2.0\ \mathrm{kg}$ over a $1.0\ \mathrm{kg}$ disk gives $a = (4.0 - 2.0)(9.8)/(4.0 + 2.0 + 0.5) = 19.6/6.5 = 3.02\ \mathrm{m/s^2}$ in one line.

If the pulley is stepped, each cord just carries its own radius into both the torque and the constraint, $a_i = \alpha r_i$. For $I = 0.40\ \mathrm{kg\,m^2}$, a $5.0\ \mathrm{kg}$ mass at $r_1 = 0.30\ \mathrm{m}$ and an $8.0\ \mathrm{kg}$ mass on the opposite side at $r_2 = 0.15\ \mathrm{m}$, eliminating the tensions the same way gives

$\alpha = \frac{m_1 g r_1 - m_2 g r_2}{I + m_1 r_1^2 + m_2 r_2^2} = \frac{(5.0)(9.8)(0.30) - (8.0)(9.8)(0.15)}{0.40 + (5.0)(0.09) + (8.0)(0.0225)}$

$\alpha = \frac{14.7 - 11.76}{0.40 + 0.45 + 0.18} = \frac{2.94}{1.03} = 2.85\ \mathrm{rad/s^2}$

The 8 kg mass is heavier but sits on the small radius, so the 5 kg mass wins. Each $m_i r_i^2$ is that block's contribution to the system's effective moment of inertia, exactly like the $\tfrac12 M$ before.

Worked example 3: the falling rod

A uniform rod of mass $M$ and length $L$, hinged at one end, is released from rest held horizontal. This is where torque stops being constant.

Gravity acts on every bit of the rod, but for torque purposes the whole weight $Mg$ acts at the centre of mass, $L/2$ from the hinge. Held horizontal, the weight is perpendicular to the rod, so the torque about the hinge is $Mg(L/2)$, and a rod about its end has $I = \tfrac13 ML^2$. So

$\alpha = \frac{\tau}{I} = \frac{MgL/2}{ML^2/3} = \frac{3g}{2L}$

For $L = 1.2\ \mathrm{m}$, that is $\alpha = 3(9.8)/(2)(1.2) = 12.25\ \mathrm{rad/s^2}$.

Two things here. The mass cancelled, as it always does when gravity is the only thing acting. And the centre of mass accelerates at $a_{cm} = \alpha(L/2) = 3g/4 = 7.35\ \mathrm{m/s^2}$, which is less than $g$. Something must be holding the rod up, and it is the hinge. Apply $F = ma$ to the rod as a whole, taking down as positive:

$Mg - F_{\text{hinge}} = M a_{cm} = M\tfrac34 g \quad\Longrightarrow\quad F_{\text{hinge}} = \tfrac14 Mg$

For $M = 2.0\ \mathrm{kg}$ that is $(0.25)(2.0)(9.8) = 4.9\ \mathrm{N}$ upward. Note what did not happen: the hinge does not support the full weight. A rod being released is not a rod in equilibrium.

Now the harder question. Once the rod starts swinging, $\theta$ changes, so the lever arm $(L/2)\cos\theta$ changes, so $\alpha$ changes every instant. You cannot use constant-$\alpha$ kinematics. Use energy instead. Take a rod held vertically on a frictionless pivot at its base, released, and ask for the speed of its tip when it hits the ground. The centre of mass drops from $L/2$ to zero, releasing $Mg(L/2)$ of potential energy, and all of it becomes rotational kinetic energy about the pivot:

$Mg\frac{L}{2} = \tfrac12 I\omega^2 = \tfrac12\left(\tfrac13 ML^2\right)\omega^2 = \tfrac16 ML^2\omega^2$

$M$ cancels. Solve for $\omega$:

$\omega = \sqrt{\frac{3g}{L}}$

With $L = 1.2\ \mathrm{m}$ and $g = 10\ \mathrm{m/s^2}$, $\omega = \sqrt{30/1.2} = \sqrt{25} = 5.0\ \mathrm{rad/s}$. The tip is at radius $L$, so its speed is $v = \omega L = (5.0)(1.2) = 6.0\ \mathrm{m/s}$. A rod released from $30^\circ$ above the horizontal instead falls through a height $(L/2)\sin 30^\circ$ before reaching horizontal, so $\omega = \sqrt{3g\sin 30^\circ/L} = \sqrt{3(9.8)(0.5)/1.4} = 3.24\ \mathrm{rad/s}$.

Rolling, and choosing a smarter pivot

Rolling without slipping is the same constraint as the cord on the pulley, $v = \omega R$ and $a = \alpha R$, but now the whole body moves. Energy conservation for something rolling from a height $h$ reads

$mgh = \tfrac12 mv^2 + \tfrac12 I\omega^2 = \tfrac12 mv^2\left(1 + \frac{I}{mR^2}\right)$

using $\omega = v/R$. Only the dimensionless ratio $I/mR^2$ survives, so mass and radius do not matter in a rolling race. A solid sphere has $I/mR^2 = 2/5$, so from $h = 2\ \mathrm{m}$ with $g = 10\ \mathrm{m/s^2}$, $v = \sqrt{2(10)(2)/1.4} = \sqrt{28.6} = 5.35\ \mathrm{m/s}$. A disk (ratio $1/2$) is slower, a hoop (ratio 1) slower still.

For rolling dynamics, you get a large shortcut by taking torques about the contact point instead of the centre. The contact point is instantaneously at rest, friction acts there, and a force applied at your pivot has zero lever arm and so disappears from the equation. Take a spool of mass $3.0\ \mathrm{kg}$, radius $R = 0.20\ \mathrm{m}$, $I_{cm} = \tfrac12 MR^2$, with a cord pulled horizontally with $12\ \mathrm{N}$ from the top of the spool. About the contact point, the cord's lever arm is $2R$ (it acts a full diameter above), and by the parallel axis theorem $I_{\text{contact}} = \tfrac12 MR^2 + MR^2 = \tfrac32 MR^2$. So

$\alpha = \frac{F(2R)}{\tfrac32 MR^2} = \frac{4F}{3MR}, \qquad a = \alpha R = \frac{4F}{3M} = \frac{4(12)}{3(3.0)} = 5.33\ \mathrm{m/s^2}$

No need to ever name the friction force. That same trick (pivot where the unknown force acts) handles the centre of percussion. Hit a hinged rod of length $L$ perpendicular with an impulse $J$ at distance $d$ from the hinge. If the hinge feels nothing, then $J$ alone gives the rod its motion, so $J = M v_{cm}$ and $Jd = I\omega = \tfrac13 ML^2\omega$. The hinge is fixed, so $v_{cm} = \omega(L/2)$. Divide the second equation by the first:

$d = \frac{\tfrac13 ML^2\omega}{M\omega L/2} = \frac{2L}{3}$

For $L = 1.0\ \mathrm{m}$, $d = 0.667\ \mathrm{m}$. That is the sweet spot on a bat, and it is the last thing $\tau = I\alpha$ gives you before you meet impulse properly in angular momentum.

Recognising the technique

Reach for $\tau = I\alpha$ when a problem gives you a moment of inertia, or gives you enough to compute one (a mass and a shape name: disk, hoop, rod, sphere), and then asks for a rate of change: an angular acceleration, a time to stop, a linear acceleration in a system that includes something rotating.

Three signals in particular:

• The problem tells you a pulley's mass, radius, or $I$. That detail is never decoration. It means two different tensions and an extra $I/R^2$ of effective mass. Write $F = ma$ for each block, $\tau = I\alpha$ for the pulley, and $a = \alpha R$ to link them, then add to kill the tensions.
• Something pivots or hinges. Torque about the hinge, $I$ about the hinge, and remember the hinge force never appears because its lever arm is zero. If the question asks for the hinge force, get $\alpha$ first, then $a_{cm} = \alpha r_{cm}$, then $F = Ma_{cm}$ for the whole body.
• Something rolls. Use $a = \alpha R$, and take torques about the contact point if an unknown friction force is in the way.

And the fork that decides your method: if the question asks for an acceleration, a force, or a time, use $\tau = I\alpha$. If it asks for a speed after moving through some angle or height, and the torque changes along the way, use energy: $Mg\Delta h = \tfrac12 I\omega^2$. Problems about falling rods and rolling balls usually want both, in that order: $\tau = I\alpha$ for the instant of release, energy for the finish.

What $\tau = I\alpha$ cannot do is handle a collision, where forces are huge and brief and $I$ itself may change. For those you need the rotational analogue of momentum, which is next: angular momentum.