A.4 Rigid body mechanics

Torque is a turning effect

The torque of a force about an axis is

$$\tau=Fr\sin\theta$$

where $r$ is the distance from the axis to the point of application and $\theta$ is the angle between $\vec r$ and $\vec F$.

Use the perpendicular lever arm

Equivalently, torque equals force multiplied by the perpendicular distance from the axis to the force’s line of action.

Choose a rotation sign

Clockwise and anticlockwise torques have opposite signs. Add torques about the specified axis rather than adding their magnitudes blindly.

Worked example from local Question Bank row 37867

Two forces produce the same rotational sense: $50\,\mathrm{N}$ at a perpendicular distance $0.50\,\mathrm{m}$ and $40\,\mathrm{N}$ at $0.20\,\mathrm{m}$.

$$\tau_{net}=(50)(0.50)+(40)(0.20)=25+8=33\,\mathrm{N\,m}\approx30\,\mathrm{N\,m}$$

If one force acted in the opposite sense, its torque would enter with the opposite sign.

Common trap

A force through the axis has zero torque, even if its magnitude is large.

Equilibrium condition

A rigid body is in rotational equilibrium when the resultant torque about any chosen axis is zero:

$$\sum\tau=0$$

Balance clockwise and anticlockwise effects

Choose an axis, assign signs, and set the sum of clockwise torques equal to the sum of anticlockwise torques. A body can still have translational equilibrium as a separate condition.

Common trap

Zero resultant torque means no angular acceleration; it does not by itself prove that the net force is zero.

Rotational second law

A non-zero resultant torque causes angular acceleration:

$$\sum\tau=I\alpha$$

Use the chosen axis

Calculate signed torques about the specified axis and use the moment of inertia about that same axis. The direction of $\alpha$ follows the resultant torque.

Common trap

Do not use translational $F=ma$ for a purely rotational equation or mix an inertia about one axis with torque about another.

Three angular quantities

Angular displacement $\theta$ describes change in orientation, angular velocity $\omega=d\theta/dt$ describes how fast orientation changes, and angular acceleration $\alpha=d\omega/dt$ describes how angular velocity changes.

Link to linear motion

At radius $r$, tangential speed is $v=r\omega$. Keep angular quantities in radians when using these relationships.

Common trap

Angular velocity is not automatically the same as linear speed; the radius is needed to connect them.

Uniform angular acceleration

When $\alpha$ is constant, use rotational SUVAT:

$$\omega=\omega_0+\alpha t,\quad \theta=\omega_0t+\frac12\alpha t^2,\quad \omega^2=\omega_0^2+2\alpha\theta$$

Choose the equation

List $\theta,\omega_0,\omega,\alpha,t$, convert revolutions to radians, and choose the equation containing the required unknown and known quantities.

Worked example from local Question Bank row 39408

A bar starts from rest and turns through six revolutions with constant $\alpha=0.110\,\mathrm{rad\,s^{-2}}$. Convert $\Delta\theta=6(2\pi)=12\pi\,\mathrm{rad}$, then

$$\omega_f^2=\omega_i^2+2\alpha\Delta\theta=0+2(0.110)(12\pi)$$
$$\omega_f=2.88\,\mathrm{rad\,s^{-1}}\approx2.9\,\mathrm{rad\,s^{-1}}$$

The equation is valid because angular acceleration is constant.

Common trap

Do not use rotational SUVAT when angular acceleration varies, and do not insert degrees or revolutions where radians are required.

Rotational inertia

Moment of inertia measures resistance to angular acceleration about an axis. It depends on total mass and how far that mass is distributed from the axis.

Compare distributions

For the same mass and outer radius, more mass farther from the axis gives larger $I$. A ring therefore has greater rotational inertia than a disk of the same mass and radius.

Common trap

Moment of inertia is not determined by mass alone; always specify the rotation axis and distribution.

Point-mass model

For discrete masses rotating about an axis,

$$I=\sum mr^2$$

where each $r$ is the perpendicular distance from the axis.

Build the sum

Treat each small sphere, blade or mass element separately, calculate $mr^2$, and add the contributions. Use symmetry when identical masses have equal radii.

Worked example from local Question Bank row 128743

Two $10\,\mathrm{kg}$ point masses are $8.0\,\mathrm{m}$ apart and rotate about the midpoint. Each is $4.0\,\mathrm{m}$ from the axis:

$$I=\sum mr^2=2(10)(4.0)^2=320\,\mathrm{kg\,m^2}$$

The full $8.0\,\mathrm{m}$ separation is not the radius of either mass.

Common trap

Do not use the distance between two masses as $r$ for both; use each mass’s distance to the rotation axis.

Torque–inertia relation

For rotation about a fixed axis,

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

Connect translation and rotation

When a force drives a rotating body or pulley, write both the translational force balance and the rotational torque balance if the system has translating and rotating parts.

Worked example from local Question Bank row 31942

A $50\,\mathrm{N}$ tangential force acts $2.0\,\mathrm{m}$ from the axis of a system with $I=450\,\mathrm{kg\,m^2}$.

$$\tau=Fr=(50)(2.0)=100\,\mathrm{N\,m}$$
$$\alpha=\frac{\tau}{I}=\frac{100}{450}=0.22\,\mathrm{rad\,s^{-2}}$$

The acceleration direction follows the signed resultant torque.

Common trap

Do not treat torque as force or use a moment of inertia that does not match the rotation axis.

Rotational momentum

For a rigid body rotating about a fixed axis,

$$L=I\omega$$

Angular momentum is directed along the rotation axis by the right-hand convention.

Use the matching inertia

The moment of inertia must be calculated about the same axis used for $L$. A larger $I$ at the same angular speed means larger angular momentum.

Worked example from local Question Bank row 31945

For $I=450\,\mathrm{kg\,m^2}$ and $\omega=1.66\,\mathrm{rad\,s^{-1}}$,

$$L=I\omega=(450)(1.66)=7.47\times10^2\,\mathrm{kg\,m^2\,s^{-1}}\approx750\,\mathrm{kg\,m^2\,s^{-1}}$$

The sign or axis direction must match the chosen rotational convention.

Common trap

Do not substitute translational momentum $mv$ for angular momentum when the question describes rotation.

Conservation condition

Angular momentum remains constant when the resultant external torque about the chosen axis is zero:

$$L_i=L_f$$

Redistribute the mass

When a skater pulls their arms inward, $I$ decreases. With angular momentum conserved, $\omega$ increases.

Worked example from local Question Bank row 34033

A $0.200\,\mathrm{kg}$ particle moving at $12.0\,\mathrm{m\,s^{-1}}$ strikes $0.60\,\mathrm{m}$ from an axis. Its initial angular momentum is

$$L_i=mvr=(0.200)(12.0)(0.60)=1.44\,\mathrm{kg\,m^2\,s^{-1}}$$

If the combined system has $I_f=0.252\,\mathrm{kg\,m^2}$ and external torque is negligible,

$$I_f\omega_f=L_i\Rightarrow\omega_f=\frac{1.44}{0.252}=5.71\,\mathrm{rad\,s^{-1}}$$

Common trap

Angular momentum conservation does not require rotational kinetic energy to remain constant when the moment of inertia changes.

Angular impulse

A torque acting for a time changes angular momentum:

$$\Delta L=\tau\Delta t=\Delta(I\omega)$$

Area under a torque–time graph

If torque varies, the signed area under a $\tau$-against-time graph gives angular impulse and therefore the change in angular momentum.

Common trap

Angular impulse has units N m s, not N s; keep it distinct from linear impulse.

Rotational energy

For a rigid body rotating about a fixed axis,

$$E_k=\frac12I\omega^2=\frac{L^2}{2I}$$

Combine forms of motion

A rolling object may have translational kinetic energy of its centre of mass and rotational kinetic energy about its centre. Include both when accounting for total kinetic energy.

Worked example from local Question Bank row 31644

A rod of weight $36.0\,\mathrm{N}$ lowers its centre of mass by $5.00/2=2.50\,\mathrm{m}$ and has $I=30.6\,\mathrm{kg\,m^2}$. If the gravitational transfer becomes rotational kinetic energy,

$$E_k=(36.0)(2.50)=90.0\,\mathrm{J}$$
$$90.0=\frac12(30.6)\omega^2\Rightarrow\omega=2.43\,\mathrm{rad\,s^{-1}}$$

The energy result is in joules; angular speed is in radians per second.

Common trap

Do not use $\frac12mv^2$ alone for a rotating body when its rotational motion contributes energy.