Rotational Kinetic Energy and Rolling
Jayden holds a solid disk and a thin hoop at the top of a ramp, same mass, same radius, and lets go at the same moment. Both roll without slipping. Which one gets to the bottom first? Both lose the same gravitational potential energy, so you might expect a tie. By the end of this page you will see why it is not.
Energy of a spinning object
A spinning flywheel is not moving from place to place, yet it can do work: it can turn a generator or keep a machine going. So it must store kinetic energy.
Think of the wheel as many small pieces of mass $m_i$ at distances $r_i$ from the axis. Each piece moves at speed $v_i = r_i\omega$, so its kinetic energy is $\tfrac{1}{2}m_i v_i^2 = \tfrac{1}{2}m_i r_i^2\omega^2$. Add up all the pieces. Because $\omega$ is the same for every piece, it comes out of the sum:
$$K_{rot} = \tfrac{1}{2}\left(\sum m_i r_i^2\right)\omega^2 = \tfrac{1}{2}I\omega^2$$Rotational inertia depends on how the mass is distributed. For the same mass and radius, a hoop (all mass at the rim, $I = MR^2$) stores more energy at a given $\omega$ than a solid disk ($I = \tfrac{1}{2}MR^2$).
A flywheel has a rotational inertia of $0.40$ kg$\cdot$m$^2$ and spins at $30$ rad/s. Calculate its rotational kinetic energy.
A flywheel has a rotational inertia of $0.40$ kg$\cdot$m$^2$ and spins at $30$ rad/s. Calculate its rotational kinetic energy.
Rolling without slipping
A tire rolling along the road turns and moves forward at the same time. If it does not slip, the point touching the ground is momentarily at rest. In one full turn the center advances one circumference, $2\pi R$. So the speed of the center of mass and the angular velocity are tied together:
This is just $v = r\omega$ applied at the rim, seen from the center of the wheel.
A wheel of radius $0.25$ m rolls without slipping. Write the relationship between the speed $v$ of its center of mass and its angular velocity $\omega$. Then find…
A wheel of radius $0.25$ m rolls without slipping. Write the relationship between the speed $v$ of its center of mass and its angular velocity $\omega$. Then find $\omega$ when $v = 5.0$ m/s.
A solid sphere of mass $2.0$ kg rolls without slipping on a flat floor with a center-of-mass speed of $3.0$ m/s. Identify the two kinds of kinetic energy the sphere has,…
A solid sphere of mass $2.0$ kg rolls without slipping on a flat floor with a center-of-mass speed of $3.0$ m/s. Identify the two kinds of kinetic energy the sphere has, and calculate its translational kinetic energy.
Total kinetic energy of a rolling object
A rolling object does two things at once, so it has two kinds of kinetic energy: translational energy of its center of mass and rotational energy about the center of mass.
The factor $I/(MR^2)$ tells you how much extra energy goes into spinning:
| Object | $I$ | Rotational share of $K_{total}$ |
|---|---|---|
| Hoop | $MR^2$ | $1/2$ |
| Solid disk | $\tfrac{1}{2}MR^2$ | $1/3$ |
| Solid sphere | $\tfrac{2}{5}MR^2$ | $2/7$ |
A hoop of mass $1.5$ kg and radius $0.40$ m rolls without slipping with a center-of-mass speed of $4.0$ m/s. For a hoop, $I = MR^2$. Calculate (a) the translational…
A hoop of mass $1.5$ kg and radius $0.40$ m rolls without slipping with a center-of-mass speed of $4.0$ m/s. For a hoop, $I = MR^2$. Calculate (a) the translational kinetic energy, (b) the rotational kinetic energy, and (c) the total kinetic energy.
Energy conservation and ramps
When an object rolls without slipping, static friction at the contact point does no work, because that point is not sliding. So mechanical energy is conserved. Choose the system as object plus Earth:
$$K_i + U_i = K_f + U_f$$For an object released from rest and rolling down a drop of height $h$:
$$Mgh = \tfrac{1}{2}Mv^2 + \tfrac{1}{2}I\omega^2 = \tfrac{1}{2}Mv^2\left(1 + \tfrac{I}{MR^2}\right)$$Solving gives
$$v = \sqrt{\dfrac{2gh}{1 + I/(MR^2)}}$$Notice that $M$ and $R$ cancel. A sliding block on a frictionless ramp has no rotational term, so $v = \sqrt{2gh}$, which is the largest possible.
Now Jayden's race. For a disk, $v = \sqrt{4gh/3} \approx 1.15\sqrt{gh}$. For a hoop, $Mgh = Mv^2$, so $v = \sqrt{gh}$. The disk is faster at the bottom and gets there first. The hoop puts half its energy into spinning, so less is left for moving down the ramp. Mass and radius do not matter; only the shape (the ratio $I/MR^2$) does. The race is answered.
A uniform solid disk of mass $3.0$ kg and radius $0.20$ m, with $I = \tfrac{1}{2}MR^2$, is released from rest and rolls without slipping down a ramp, dropping a vertical…
A uniform solid disk of mass $3.0$ kg and radius $0.20$ m, with $I = \tfrac{1}{2}MR^2$, is released from rest and rolls without slipping down a ramp, dropping a vertical height of $0.60$ m. Use energy conservation with $g = 9.8$ m/s$^2$ to find the speed of its center of mass at the bottom.
Jayden releases a solid disk ($I = \tfrac{1}{2}MR^2$) and a thin hoop ($I = MR^2$) from rest at the top of the same ramp, a vertical height of $h = 0.80$ m above the…
Jayden releases a solid disk ($I = \tfrac{1}{2}MR^2$) and a thin hoop ($I = MR^2$) from rest at the top of the same ramp, a vertical height of $h = 0.80$ m above the bottom. Both have the same mass and radius and roll without slipping. (a) Derive an expression for each object's center-of-mass speed at the bottom in terms of $g$ and $h$, and calculate both values with $g = 9.8$ m/s$^2$. (b) Determine which object reaches the bottom first and explain why, in terms of how the energy is divided between translation and rotation. (c) State whether the result depends on the mass or radius of the objects.
A uniform solid sphere ($I = \tfrac{2}{5}MR^2$) of mass $0.50$ kg and radius $0.05$ m rolls without slipping along a level track at $2.0$ m/s and then rolls up a ramp.…
A uniform solid sphere ($I = \tfrac{2}{5}MR^2$) of mass $0.50$ kg and radius $0.05$ m rolls without slipping along a level track at $2.0$ m/s and then rolls up a ramp. Assume no energy is lost. (a) Find the sphere's total kinetic energy on the level track and the fraction that is rotational. (b) Find the maximum vertical height the sphere reaches on the ramp, using $g = 9.8$ m/s$^2$. (c) Compare this height with the height a block sliding on a frictionless ramp with the same initial speed would reach, and explain the difference.
Work and power in rotation
The work-energy theorem works for spinning objects too. If a motor is the only thing doing work on a flywheel and nothing is lost to friction, then
$$W_{net} = \Delta K_{rot} = \tfrac{1}{2}I\omega_f^2 - \tfrac{1}{2}I\omega_i^2$$Power is still energy transferred per unit time: $P_{avg} = W/\Delta t$.
A motor spins a flywheel of rotational inertia $0.80$ kg$\cdot$m$^2$ from rest to $50$ rad/s. (a) Find the work the motor does on the flywheel, assuming no friction…
A motor spins a flywheel of rotational inertia $0.80$ kg$\cdot$m$^2$ from rest to $50$ rad/s. (a) Find the work the motor does on the flywheel, assuming no friction losses. (b) If this took $4.0$ s, find the average power delivered.