A.4 Rigid body mechanics

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.

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.

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.

Link to linear motion

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

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.

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.

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.

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.

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.

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,

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

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.

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.