Free AP Physics 1 Energy and Momentum of Rotating Systems

AP Physics 1 · Energy and Momentum of Rotating Systems · free lesson

Rotational Kinetic Energy

The idea, then worked examples solved step by step, then practice questions whose numbers change every time you try them.

The idea

The energy of a spinning object

A spinning bicycle wheel can hurt a hand that is put in its spokes, even though the wheel is staying where it is. A flywheel can store enough energy to run a machine for a while. A wheel in place has no motion of its center, so the energy must come from something else: every part of the wheel is moving in a circle, and every moving part has kinetic energy.

Definition.

The rotational kinetic energy of a rigid system is the energy it has because it rotates. For a rotational inertia II about its axis and an angular velocity ω\omega about that axis,

Krot=12 Iω2\begin{aligned} K_{rot} &= \tfrac{1}{2}\, I \omega^2 \end{aligned}

It is measured in joules (J), like every energy. The angular velocity must be in rad/s.

A flywheel with a rotational inertia of 0.250 kg⋅m20.250\,\mathrm{kg{\cdot}m^{2}} turns at 12.0 rad/s12.0\,\mathrm{rad/s}. Square the angular velocity, multiply by the rotational inertia, and take half.

K=Iω22=0.25 kg⋅m2×(12 rad/s)22=18.0 J\begin{aligned} K &= \frac{I\omega^{2}}{2} \\ &= \frac{0.25\,\mathrm{kg{\cdot}m^{2}} \times \left(12\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 18.0\,\mathrm{J} \end{aligned}

The rotational kinetic energy is 18.0 J18.0\,\mathrm{J}. The formula looks like 12mv2\tfrac12 mv^2 with II in the place of the mass and ω\omega in the place of the speed. That is no accident: II is the rotational counterpart of mass.

Why the formula has this form

A rigid system is made of many small parts. Each part of mass mim_i moves in a circle of radius rir_i about the axis, and all the parts share the same angular velocity ω\omega. A part at distance rir_i has speed vi=riωv_i = r_i\omega and kinetic energy 12mivi2\tfrac12 m_i v_i^2. Adding the energies of all the parts gives

Krot=∑12mivi2=∑12miri2ω2=12(∑miri2)ω2=12Iω2\begin{aligned} K_{rot} &= \sum \tfrac{1}{2} m_i v_i^2 = \sum \tfrac{1}{2} m_i r_i^2 \omega^2 = \tfrac{1}{2} \left(\sum m_i r_i^2\right) \omega^2 = \tfrac{1}{2} I \omega^2 \end{aligned}

Here is a small example. Two equal masses sit at the ends of a light rod that spins about its middle.

Each mass is 0.400 m0.400\,\mathrm{m} from the axis, so at 6.00 rad/s6.00\,\mathrm{rad/s} its speed is 2.40 m/s2.40\,\mathrm{m/s}, and the two arrows point opposite ways. Add the kinetic energies of the two masses, or use the rotational inertia of the whole system. Both give the same energy.

K1=mv22=0.5 kg×(2.4 m/s)22=1.44 JKt=2K1=2×1.44 J=2.88 JI=2mr2=2×0.5 kg×(0.4 m)2=0.160 kg⋅m2Kr=Iω22=0.16 kg⋅m2×(6 rad/s)22=2.88 J\begin{aligned} K_1 &= \frac{mv^{2}}{2} \\ &= \frac{0.5\,\mathrm{kg} \times \left(2.4\,\mathrm{m/s}\right)^{2}}{2} \\ &= 1.44\,\mathrm{J} \\ K_t &= 2K_1 \\ &= 2 \times 1.44\,\mathrm{J} \\ &= 2.88\,\mathrm{J} \\ I &= 2mr^{2} \\ &= 2 \times 0.5\,\mathrm{kg} \times \left(0.4\,\mathrm{m}\right)^{2} \\ &= 0.160\,\mathrm{kg{\cdot}m^{2}} \\ K_r &= \frac{I\omega^{2}}{2} \\ &= \frac{0.16\,\mathrm{kg{\cdot}m^{2}} \times \left(6\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 2.88\,\mathrm{J} \end{aligned}

Each mass has 1.44 J1.44\,\mathrm{J}, so together 2.88 J2.88\,\mathrm{J}. The rotational inertia is 0.160 kg⋅m20.160\,\mathrm{kg{\cdot}m^{2}} and 12Iω2\tfrac12 I\omega^2 is 2.88 J2.88\,\mathrm{J}. The two ways agree, because 12Iω2\tfrac12 I\omega^2 is just the sum of the 12mv2\tfrac12 mv^2 of all the parts.

Rotational kinetic energy is the kinetic energy of the moving parts

The rod above does not go anywhere: its center is at rest. Yet it has energy, because each point in the system moves, and so has linear speed and kinetic energy. A rigid system can have rotational kinetic energy while its center of mass is at rest. There is no separate kind of energy hiding in a rotation: the rotational kinetic energy is the total kinetic energy of the parts, and 12Iω2\tfrac12 I\omega^2 is a convenient way to add it up.

Rule.

For an object rotating about a fixed axis, the rotational kinetic energy 12Iω2\tfrac12 I\omega^2 is its total kinetic energy. It is equal to the sum of the translational kinetic energies 12mivi2\tfrac12 m_iv_i^2 of all its points.

For a single small object of mass mm going around an axis at distance rr, we have I=mr2I = mr^{2} and v=rωv = r\omega. Then 12Iω2=12mr2ω2=12mv2\tfrac12 I\omega^2 = \tfrac12 mr^2\omega^2 = \tfrac12 mv^2: the two expressions are the same number.

Ka=mv22=0.3 kg×(4 m/s)22=2.40 JI=mr2=0.3 kg×(0.5 m)2=0.0750 kg⋅m2Kb=Iω22=0.075 kg⋅m2×(8 rad/s)22=2.40 J\begin{aligned} K_a &= \frac{mv^{2}}{2} \\ &= \frac{0.3\,\mathrm{kg} \times \left(4\,\mathrm{m/s}\right)^{2}}{2} \\ &= 2.40\,\mathrm{J} \\ I &= mr^{2} \\ &= 0.3\,\mathrm{kg} \times \left(0.5\,\mathrm{m}\right)^{2} \\ &= 0.0750\,\mathrm{kg{\cdot}m^{2}} \\ K_b &= \frac{I\omega^{2}}{2} \\ &= \frac{0.075\,\mathrm{kg{\cdot}m^{2}} \times \left(8\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 2.40\,\mathrm{J} \end{aligned}

A 0.300 kg0.300\,\mathrm{kg} ball on a light string, 0.500 m0.500\,\mathrm{m} from the axis, turns at 8.00 rad/s8.00\,\mathrm{rad/s}, so its speed is 4.00 m/s4.00\,\mathrm{m/s}. As a moving object it has 2.40 J2.40\,\mathrm{J}; as a rotating system with II = 0.0750 kg⋅m20.0750\,\mathrm{kg{\cdot}m^{2}} it has 2.40 J2.40\,\mathrm{J}.

How the energy depends on II and ω\omega

Because ω\omega is squared, the energy grows quickly with the rate of spin. Move the slider to change the angular velocity of a disk. The arrows show the speed of three points; the point twice as far from the axis moves twice as fast. The bar shows the rotational kinetic energy on a scale that is the same each time.

From 4 rad/s to 8 rad/s the bar becomes four times as long. Make the rotational inertia twice as large instead and the energy only doubles. The graph shows the energy against the angular velocity for a system and for the same system with twice the rotational inertia.

The blue curve is for 0.400 kg⋅m20.400\,\mathrm{kg{\cdot}m^{2}} and the red one for 0.800 kg⋅m20.800\,\mathrm{kg{\cdot}m^{2}}. At 4.00 rad/s4.00\,\mathrm{rad/s} the blue system has 3.20 J3.20\,\mathrm{J}; at 8.00 rad/s8.00\,\mathrm{rad/s} it has 12.8 J12.8\,\mathrm{J}. Both are parabolas through the origin, and at each angular velocity the red curve is twice as high.

Rule.

Scaling. If the rotational inertia is multiplied by a factor aa and the angular velocity by a factor bb, the rotational kinetic energy is multiplied by ab2ab^{2}.

Moving and spinning together

Many objects turn while their centers also move: a baton thrown in the air, a tumbling gymnast, a spinning ball in flight. The kinetic energy of such a rigid system is the sum of two parts, one for the motion of its center of mass and one for the rotation about the center of mass.

Rule.

The total kinetic energy of a rigid system is the sum of its translational kinetic energy, due to the linear motion of its center of mass, and its rotational kinetic energy, due to its rotation about its center of mass:

K=12 Mvcm2+12 Icm ω2\begin{aligned} K &= \tfrac{1}{2}\, M v_{cm}^2 + \tfrac{1}{2}\, I_{cm}\, \omega^2 \end{aligned}

A baton of mass 0.400 kg0.400\,\mathrm{kg} flies through the air with its center of mass moving at 5.00 m/s5.00\,\mathrm{m/s} while it spins about that center at 15.0 rad/s15.0\,\mathrm{rad/s}. About its center its rotational inertia is 0.0200 kg⋅m20.0200\,\mathrm{kg{\cdot}m^{2}}.

Ktr=Mvc22=0.4 kg×(5 m/s)22=5.00 JKro=Icω22=0.02 kg⋅m2×(15 rad/s)22=2.25 JKtot=Ktr+Kro=5 J+2.25 J=7.25 J\begin{aligned} K_{tr} &= \frac{Mv_c^{2}}{2} \\ &= \frac{0.4\,\mathrm{kg} \times \left(5\,\mathrm{m/s}\right)^{2}}{2} \\ &= 5.00\,\mathrm{J} \\ K_{ro} &= \frac{I_c\omega^{2}}{2} \\ &= \frac{0.02\,\mathrm{kg{\cdot}m^{2}} \times \left(15\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 2.25\,\mathrm{J} \\ K_{tot} &= K_{tr} + K_{ro} \\ &= 5\,\mathrm{J} + 2.25\,\mathrm{J} \\ &= 7.25\,\mathrm{J} \end{aligned}

The center's motion carries 5.00 J5.00\,\mathrm{J}, the spin carries 2.25 J2.25\,\mathrm{J}, and the total is 7.25 J7.25\,\mathrm{J}. Energies are added as plain numbers, so the two parts can simply be added. Notice that neither part alone is the whole story: with only the translational part you would miss 2.25 J2.25\,\mathrm{J}.

A scalar

Rotational kinetic energy is a scalar. The formula has ω2\omega^2 in it, so it does not care whether the system turns clockwise or counterclockwise, and it is never negative. A wheel at +6+6 rad/s and a wheel at −6-6 rad/s with the same II have the same rotational kinetic energy. The rotational kinetic energies of several parts add as plain numbers.

Warning.

Three traps. The angular velocity must be in rad/s: for a wheel at 60 rpm use ω=2π\omega = 2\pi rad/s, not 60. Doubling the angular velocity quadruples the energy. And the sign of ω\omega (clockwise or counterclockwise) is a direction; it does not make the energy negative.

Worked examples

Worked examples

Six examples of the kinds the AP exam asks on this topic: rotational kinetic energy from the rotational inertia and the angular velocity, a "what changes if", the sum over the parts of a system, a system that moves and spins, finding an angular velocity from an energy, and a claim justified.

Example.

Example 1. A spinning fan

The blades of a ceiling fan have a rotational inertia of 0.0450 kg⋅m20.0450\,\mathrm{kg{\cdot}m^{2}} about the shaft. The fan turns at 300 rpm300\,\mathrm{rpm}. What is the rotational kinetic energy of the blades?

Solution

The angular velocity must be in rad/s. The fan makes 300300 turns in 60 s, which is 300300/60 = 5.005.00 turns per second. Each turn is 2π2\pi rad, so we write the 2π2\pi ourselves.

ω=2πf=2π×5 Hz=31.4 rad/s\begin{aligned} \omega &= 2\pi f \\ &= 2\pi \times 5\,\mathrm{Hz} \\ &= 31.4\,\mathrm{rad/s} \end{aligned}
K=Iω22=0.045 kg⋅m2×(31.4 rad/s)22=22.2 J\begin{aligned} K &= \frac{I\omega^{2}}{2} \\ &= \frac{0.045\,\mathrm{kg{\cdot}m^{2}} \times \left(31.4\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 22.2\,\mathrm{J} \end{aligned}
Result.

The blades turn at 31.4 rad/s31.4\,\mathrm{rad/s} and have a rotational kinetic energy of 22.2 J22.2\,\mathrm{J}. Using 300 in place of the angular velocity would give an energy about a hundred times too big.

Example.

Example 2. What changes if

A turntable has a rotational inertia of 0.800 kg⋅m20.800\,\mathrm{kg{\cdot}m^{2}} and turns at 5.00 rad/s5.00\,\mathrm{rad/s}. A second turntable has a rotational inertia three times as large and turns at half that angular velocity. Which has more rotational kinetic energy, and how many times as much?

Solution

Use the scaling rule: the energy is multiplied by ab2ab^{2}, where aa is the factor on II and bb the factor on ω\omega. Here a=3a = 3 and b=12b = \frac{1}{2}.

K1=I1ω22=0.8 kg⋅m2×(5 rad/s)22=10.0 JK2=I2(ω2)22=2.4 kg⋅m2×(5 rad/s2)22=7.50 J\begin{aligned} K_1 &= \frac{I_1\omega^{2}}{2} \\ &= \frac{0.8\,\mathrm{kg{\cdot}m^{2}} \times \left(5\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 10.0\,\mathrm{J} \\ K_2 &= \frac{I_2(\frac{\omega}{2})^{2}}{2} \\ &= \frac{2.4\,\mathrm{kg{\cdot}m^{2}} \times (\frac{5\,\mathrm{rad/s}}{2})^{2}}{2} \\ &= 7.50\,\mathrm{J} \end{aligned}
Result.

The first turntable has 10.0 J10.0\,\mathrm{J} and the second has 7.50 J7.50\,\mathrm{J}. The second has 0.7500.750 times as much, which is 3×(1/2)2=3/43 \times (1/2)^2 = 3/4. The larger rotational inertia is outweighed by the smaller angular velocity, because the angular velocity is squared.

Example.

Example 3. The energy part by part

Three small objects are fixed to a light, rigid frame that turns about an axis at the dot, seen from above. The masses are 1.00 kg1.00\,\mathrm{kg}, 2.00 kg2.00\,\mathrm{kg} and 0.500 kg0.500\,\mathrm{kg}, at distances of 0.300 m0.300\,\mathrm{m}, 0.200 m0.200\,\mathrm{m} and 0.600 m0.600\,\mathrm{m} from the axis. The frame turns at 4.00 rad/s4.00\,\mathrm{rad/s}. Find the rotational kinetic energy in two ways, and say which object has the most energy.

Solution

Each object moves in a circle at speed v=rωv = r\omega, so each has its own kinetic energy. The total is their sum. The other way is to find II for the whole system and use 12Iω2\tfrac12 I\omega^2.

v1=r1ω=0.3 m×4 rad/s=1.20 m/sv2=r2ω=0.2 m×4 rad/s=0.800 m/sv3=r3ω=0.6 m×4 rad/s=2.40 m/sK1=m1v122=1 kg×(1.2 m/s)22=0.720 JK2=m2v222=2 kg×(0.8 m/s)22=0.640 JK3=m3v322=0.5 kg×(2.4 m/s)22=1.44 JKt=K1+K2+K3=0.72 J+0.64 J+1.44 J=2.80 J\begin{aligned} v_1 &= r_1\omega \\ &= 0.3\,\mathrm{m} \times 4\,\mathrm{rad/s} \\ &= 1.20\,\mathrm{m/s} \\ v_2 &= r_2\omega \\ &= 0.2\,\mathrm{m} \times 4\,\mathrm{rad/s} \\ &= 0.800\,\mathrm{m/s} \\ v_3 &= r_3\omega \\ &= 0.6\,\mathrm{m} \times 4\,\mathrm{rad/s} \\ &= 2.40\,\mathrm{m/s} \\ K_1 &= \frac{m_1v_1^{2}}{2} \\ &= \frac{1\,\mathrm{kg} \times \left(1.2\,\mathrm{m/s}\right)^{2}}{2} \\ &= 0.720\,\mathrm{J} \\ K_2 &= \frac{m_2v_2^{2}}{2} \\ &= \frac{2\,\mathrm{kg} \times \left(0.8\,\mathrm{m/s}\right)^{2}}{2} \\ &= 0.640\,\mathrm{J} \\ K_3 &= \frac{m_3v_3^{2}}{2} \\ &= \frac{0.5\,\mathrm{kg} \times \left(2.4\,\mathrm{m/s}\right)^{2}}{2} \\ &= 1.44\,\mathrm{J} \\ K_t &= K_1 + K_2 + K_3 \\ &= 0.72\,\mathrm{J} + 0.64\,\mathrm{J} + 1.44\,\mathrm{J} \\ &= 2.80\,\mathrm{J} \end{aligned}
I=m1r12+m2r22+m3r32=1 kg×(0.3 m)2+2 kg×(0.2 m)2+0.5 kg×(0.6 m)2=0.350 kg⋅m2Kr=Iω22=0.35 kg⋅m2×(4 rad/s)22=2.80 J\begin{aligned} I &= m_1r_1^{2} + m_2r_2^{2} + m_3r_3^{2} \\ &= 1\,\mathrm{kg} \times \left(0.3\,\mathrm{m}\right)^{2} + 2\,\mathrm{kg} \times \left(0.2\,\mathrm{m}\right)^{2} + 0.5\,\mathrm{kg} \times \left(0.6\,\mathrm{m}\right)^{2} \\ &= 0.350\,\mathrm{kg{\cdot}m^{2}} \\ K_r &= \frac{I\omega^{2}}{2} \\ &= \frac{0.35\,\mathrm{kg{\cdot}m^{2}} \times \left(4\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 2.80\,\mathrm{J} \end{aligned}
Result.

The parts have 0.720 J0.720\,\mathrm{J}, 0.640 J0.640\,\mathrm{J} and 1.44 J1.44\,\mathrm{J}, a total of 2.80 J2.80\,\mathrm{J}. With II = 0.350 kg⋅m20.350\,\mathrm{kg{\cdot}m^{2}} the formula gives 2.80 J2.80\,\mathrm{J}, the same. The lightest object, 0.500 kg0.500\,\mathrm{kg}, has the most energy: it is the farthest out and so moves the fastest, and the energy goes as the square of the speed.

Example.

Example 4. Spinning while moving

A solid disk of mass 0.600 kg0.600\,\mathrm{kg} and radius 0.150 m0.150\,\mathrm{m} is tossed into the air. Its center moves at 3.00 m/s3.00\,\mathrm{m/s} and it spins about its center at 20.0 rad/s20.0\,\mathrm{rad/s}. Its rotational inertia about its center is 12MR2\tfrac12 MR^2. Find the total kinetic energy, and the part of it that is rotational.

Solution

The total is the translational kinetic energy of the center of mass plus the rotational kinetic energy about the center of mass.

Ic=MR22=0.6 kg×(0.15 m)22=0.00675 kg⋅m2Ktr=Mvc22=0.6 kg×(3 m/s)22=2.70 JKro=Icω22=0.00675 kg⋅m2×(20 rad/s)22=1.35 JKtot=Ktr+Kro=2.7 J+1.35 J=4.05 J\begin{aligned} I_c &= \frac{MR^{2}}{2} \\ &= \frac{0.6\,\mathrm{kg} \times \left(0.15\,\mathrm{m}\right)^{2}}{2} \\ &= 0.00675\,\mathrm{kg{\cdot}m^{2}} \\ K_{tr} &= \frac{Mv_c^{2}}{2} \\ &= \frac{0.6\,\mathrm{kg} \times \left(3\,\mathrm{m/s}\right)^{2}}{2} \\ &= 2.70\,\mathrm{J} \\ K_{ro} &= \frac{I_c\omega^{2}}{2} \\ &= \frac{0.00675\,\mathrm{kg{\cdot}m^{2}} \times \left(20\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 1.35\,\mathrm{J} \\ K_{tot} &= K_{tr} + K_{ro} \\ &= 2.7\,\mathrm{J} + 1.35\,\mathrm{J} \\ &= 4.05\,\mathrm{J} \end{aligned}
Result.

The center's motion has 2.70 J2.70\,\mathrm{J} and the spin has 1.35 J1.35\,\mathrm{J}, so the total is 4.05 J4.05\,\mathrm{J}. The rotational part is one third of the total, 33.333.3 percent.

Example.

Example 5. Finding the angular velocity

A flywheel with a rotational inertia of 2.00 kg⋅m22.00\,\mathrm{kg{\cdot}m^{2}} must store 25.0 J25.0\,\mathrm{J} of rotational kinetic energy. What angular velocity does it need?

Solution

Solve K=12Iω2K = \tfrac12 I\omega^2 for ω\omega: ω=2K/I\omega = \sqrt{2K/I}.

ω=2KI=2×25 J2 kg⋅m2=5.00 rad/s\begin{aligned} \omega &= \sqrt{\frac{2K}{I}} \\ &= \sqrt{\frac{2 \times 25\,\mathrm{J}}{2\,\mathrm{kg{\cdot}m^{2}}}} \\ &= 5.00\,\mathrm{rad/s} \end{aligned}
Result.

It needs 5.00 rad/s5.00\,\mathrm{rad/s}. That is 47.747.7 rpm, found by dividing by 2π2\pi rad per turn and multiplying by 60 s per minute.

Example.

Example 6. A claim to justify

A hoop and a solid disk have the same mass MM and the same radius RR. Each spins about an axis through its center, perpendicular to its plane, at the same angular velocity. A student says, "The two have the same rotational kinetic energy, because they have the same mass and the same angular velocity." Is the student right?

Solution

No. The rotational kinetic energy is 12Iω2\tfrac12 I\omega^2, and it depends on the rotational inertia, not on the mass alone. The rotational inertia depends on where the mass is. The hoop has all its mass at the radius RR, so Ihoop=MR2I_{hoop} = MR^2. The disk has some of its mass nearer the axis, so Idisk=12MR2I_{disk} = \tfrac12 MR^2.

Ihoop=MR2=2 kg×(0.3 m)2=0.180 kg⋅m2Idisk=MR22=2 kg×(0.3 m)22=0.0900 kg⋅m2Khoop=Ihoopω22=0.18 kg⋅m2×(10 rad/s)22=9.00 JKdisk=Idiskω22=0.09 kg⋅m2×(10 rad/s)22=4.50 J\begin{aligned} I_{hoop} &= MR^{2} \\ &= 2\,\mathrm{kg} \times \left(0.3\,\mathrm{m}\right)^{2} \\ &= 0.180\,\mathrm{kg{\cdot}m^{2}} \\ I_{disk} &= \frac{MR^{2}}{2} \\ &= \frac{2\,\mathrm{kg} \times \left(0.3\,\mathrm{m}\right)^{2}}{2} \\ &= 0.0900\,\mathrm{kg{\cdot}m^{2}} \\ K_{hoop} &= \frac{I_{hoop}\omega^{2}}{2} \\ &= \frac{0.18\,\mathrm{kg{\cdot}m^{2}} \times \left(10\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 9.00\,\mathrm{J} \\ K_{disk} &= \frac{I_{disk}\omega^{2}}{2} \\ &= \frac{0.09\,\mathrm{kg{\cdot}m^{2}} \times \left(10\,\mathrm{rad/s}\right)^{2}}{2} \\ &= 4.50\,\mathrm{J} \end{aligned}
Result.

For 2.00 kg2.00\,\mathrm{kg}, 0.300 m0.300\,\mathrm{m} and 10.0 rad/s10.0\,\mathrm{rad/s}, the hoop has 9.00 J9.00\,\mathrm{J} and the disk has 4.50 J4.50\,\mathrm{J}. The hoop has twice the rotational kinetic energy. Every point of the hoop is at the full radius, so every point moves at the greatest speed, RωR\omega. The points of the disk nearer the axis move more slowly and have less kinetic energy.

Practice

Answer AP-style questions on rotational kinetic energy. Each answer is checked, with the full solution, and every try brings new numbers. Practice needs a free account: it keeps your progress.

Practise rotational kinetic energy with new numbers