FFR - Rotational Energy

Original textbook reading.

Rotational energy and ideal rolling fit AP. Quantitative slipping-to-rolling and frictional transition calculations are extra exploration; AP treats slipping qualitatively.

Now that you know how to work with the basics behind rotating bodies, you will need to know how to combine the analysis of a rotating body with a moving one. The ‘moving’ referred to there is also described as ‘translation’, basically just moving along some path in space, instead of around a central point.

Now, when you work with problems that combine the two ideas, one of the most reliable equations is the energy balance. Recall that the energy balance as we know it is:

$K E_{1} + P E_{1} + W_{a d d e d} = K E_{2} + P E_{2} + H_{L}$ (Equation 6.5.1)

$W$ is work, $K E$ is kinetic energy, $P E$ is potential energy and $H_{L}$ is heat loss.

Rotational Kinetic Energy

Also recall that the potential energy had two components, gravitational potential energyⓘ and spring potential energy.

If you want to use this equation for a rotating object, you have to learn the energy that a rotating object has.

This energy is called Rotational Energyⓘ, or Angular Kinetic Energy. The formula for it is given as:

$K E_{r o t a t i o n a l} = \frac{1}{2} I {(\vec{ω})}^{2}$ (Equation 6.5.2)

This is very similar to the translational kinetic energy, so whenever you use the energy equation in situations that have a rotating object, you have to split the kinetic energy into the translational kinetic energy and the rotational kinetic energy.

Example Problem 6.5.1

A uniform solid disk wheel of mass 1.0 kg and diameter 0.50 m has zero center-of-mass speed and spin magnitude 50 rad/s when placed on a rough level floor. It slips until pure rolling begins and then rolls toward a 3.0 m-high hill without further losses. Does it reach the top? If not, find its maximum rise above the foot of the hill. Ignore the initial drop height.

Step 1. Identify the initial energy in the wheel

Initially, the wheel has only rotational energy, as it is spinning but not moving along the ground yet. This means that, assuming we set our height datum at the initial height, the energy will be:

$E = \frac{1}{2} I {(\vec{ω})}^{2}$

The moment of inertia for a thin, solid disk spinning as a wheel would be is:

$I = \frac{1}{2} m r^{2}$

$I = \frac{1}{2} \cdot 1 kg \cdot {(0.25 m)}^{2}$

$I = 0.0313 kg \cdot m^{2}$

This makes the energy:

Initial spin energy is K_i = ½(0.03125)(50²) = 39.0625 J. During slipping, friction accelerates the center and reduces the spin; some energy becomes internal energy.

Step 2. Find the motion after slipping ends.

Let k = I/(mR²) = 1/2. A friction impulse J gives mv = J and I(ω_0 − ω) = JR. At rolling onset, v = Rω.

v = [k/(1 + k)]Rω_0 = (1/3)(0.25)(50) = 4.16667 m/s; ω = v/R = 16.6667 rad/s.

The rolling kinetic energy is K_roll = ½mv² + ½I(v/R)² = (3/4)mv².

K_roll = (3/4)(1.0)(4.16667²) = 13.0208 J. The slipping loss is 39.0625 − 13.0208 = 26.0417 J.

The 3.0 m hill requires mgh = 29.43 J, which exceeds the 13.0208 J remaining.

The wheel cannot reach the hilltop.

Step 3. Find the maximum rise.

While rolling up the hill without slipping or further losses, the post-slip kinetic energy converts into gravitational potential energy.

mgh_max = 13.0208 J.

h_max = 13.0208/(1.0 × 9.81) = 1.32730 m.

At the maximum rise, both the center-of-mass speed and spin speed are zero in the ideal rolling model.

Conclusion: the disk rises 1.33 m above the hill’s foot.

The slipping stage conserves the appropriate impulse relation, while dissipating mechanical energy.

The later ideal rolling stage conserves mechanical energy.

The relation v = Rω applies only after rolling without slipping begins.

The source’s 3.58 m/s hilltop result relied on an impossible lossless spin-to-roll transition.

Maximum rise: h_max ≈ 1.33 m; the 3.0 m hilltop is not reached.

Pause & recall

Pop Quiz 6.5.1 Rotational Energy

Question 1 of 5 · 0 correct

AP Topic 6.2 · Read the worked solution

A uniform solid disk wheel rolls without slipping up a 15° ramp at an initial center-of-mass speed of 5.0 m/s. Its mass is 3.0 g and radius 0.50 m. With no energy losses, how far along the ramp does it travel before stopping?

Read every full question without answers

Question 1

$5.6 m$
$7.4 m$
$8.6 m$
$9.2 m$

Separate worked answer

Question 2

A uniform thin-rod propeller of length 2.0 m and mass 15 kg starts from rest and reaches 120 rpm after turning 5.0 revolutions. What constant net torque acts on it?

$12.6 N \cdot m$
$14.2 N \cdot m$
$10.5 N \cdot m$
$6.8 N \cdot m$

Separate worked answer

Question 3

A uniform solid sphere of radius 0.50 m and mass 1.0 kg starts from rest and rolls without slipping down a hill with a vertical drop of 6.0 m. With no losses, what is its center-of-mass speed at the bottom?

$4.58 m/s$
$8.50 m/s$
$9.17 m/s$
$12.7 m/s$

Separate worked answer

Question 4

A barrel modeled as a thin cylindrical shell of mass 10 kg and radius 1.0 m reaches a rough level floor with center-of-mass speed 50 m/s and zero spin. It slips until rolling without slipping begins. Neglect its drop height and rolling resistance. What is its angular speed when rolling begins?

25.0 rad/s
$45.1 rad/s$
$40.2 rad/s$
$30.5 rad/s$

Separate worked answer

Question 5

A uniform solid sphere of radius 0.25 m and mass 0.50 kg slides at 5.0 m/s with zero spin. It then enters a rough level floor and slips until rolling without slipping begins. What is its final center-of-mass speed?

$5 m/s$
$3.8 m/s$
$2.7 m/s$
3.57 m/s

Separate worked answer

Chapter 6 Equations List

Section 1:

a_c = v²/r (Equation 6.1.1), directed toward the center.

ΣF_inward = m v²/r (Equation 6.1.2).

Section 2:

$I = m r^{2}$ (Equation 6.2.1)
$I = \int_{V} ρ d^{2} d V$ (Equation 6.2.2)
$I_{z} = I_{c} + m R^{2}$ (Equation 6.2.3)

Section 3:

τ = r × F; |τ| = rF sinθ; τ_net,z = I_zα; a_t = αr; v_t = ωr. The last two are magnitudes; their vector forms use cross products. (Equations 6.3.1–6.3.5)

Section 4:

${\vec{H}}_{0} = \vec{r} \times \vec{p}$ (Equation 6.4.1)

${\vec{H}}_{0} = I \vec{ω}$ (Equation 6.4.2)

${({\vec{H}}_{0})}_{i} = {({\vec{H}}_{0})}_{f}$ (Equation 6.4.3)

$I_{i} {\vec{ω}}_{i} = I_{f} {\vec{ω}}_{f}$ (Equation 6.4.4)

Section 5:

$K E_{1} + P E_{1} + W_{a d d e d} = K E_{2} + P E_{2} + H_{L}$

(Equation 6.5.1)
$K E_{r o t a t i o n a l} = \frac{1}{2} I {(\vec{ω})}^{2}$ (Equation 6.5.2)

Chapter 6 Problems

Work it out

Original problem 1

The plane and Tarzan cases require a specified plane of motion; the answer uses a level turn for the plane and a horizontal swing around the tree for Tarzan.

Problem text

Identify the physical forces whose inward components keep these objects on circular paths: (a) a plane in a level banked turn, (b) a car on a wet banked curve, (c) Tarzan swinging around a tree on a vine, and (d) a planet orbiting the Sun. Explain each; “centripetal force” alone is not an identification.

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Work it out

Original problem 2

The prompt is explicitly reframed as a consistency check. “Tons” is interpreted as metric tonnes. The very distant speed-based calculation is only an Earth-only mathematical model; other celestial gravity would matter there.

Problem text

A shuttle of mass 40 metric tonnes is described as moving in a circular Earth orbit at 45 m/s while Earth’s gravitational force on it is 10 kN. Check whether these values describe one circular orbit. If they do not, find the altitude implied by the force and the circular speed required there. Use G = 6.67430 × 10⁻¹¹ N·m²/kg², M_E = 5.98 × 10²⁴ kg and R_E = 6.38 × 10⁶ m.

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section breakdown

Rotational Kinetic Energy
Example 6.5.1- Kinetic Energy of a Wheel
Chapter Equations
Chapter Problems

Reviewed physics notes (25)

These corrections keep the original lesson’s purpose while fixing errors in the source edition.

The original no-loss transition from spin only to rolling is not physically realizable on a passive fixed floor. — Reference 1
Retain the initial rotational-energy calculation but do not conserve it through slipping. — Reference 1
Replace the invalid no-loss landing stage. — Reference 1
Impulse and angular impulse govern the passive spin-to-roll transition. — Reference 1
The resulting speed follows from both translation and rotation, not energy conservation. — Reference 1
Use the correct energy after slipping. — Reference 1
Only one third of the initial spin energy remains for this solid-disk landing. — Reference 1
The repaired physical model reverses the source reachability conclusion. — Reference 1
It does not have enough post-slip rolling energy. — Reference 1
Compute the reachable height. — Reference 1
Separate the dissipative landing from the conservative rolling climb. — Reference 1
Both types of rolling energy become gravitational energy at the turning point. — Reference 1
Replace the invalid hilltop-speed calculation. — Reference 1
This is a turning point, not a 3 m hilltop crossing. — Reference 1
Replace the continuation of the invalid hilltop solution with the corrected conclusion. — Reference 1
Repair the repeated vector-squared equation. — Reference 1
Repair the repeated centripetal-force equation. — Reference 1
The equation list repeats scalar expressions with invalid vector notation. — Reference 1
The 7.4 m key requires a solid-disk model and no-slip rolling. — Reference 1
Initial rest and constant torque are needed for this work/kinematics question. — Reference 1
The 9.17 m/s key depends on a solid sphere and rest release. — Reference 1
The passive floor transition necessarily dissipates energy; the source lossless assumption is contradictory. — Reference 1
For I = mR², the no-spin slide-to-roll transition gives v_f = v_i/2 and ω_f = v_f/R. — Reference 1
Specify the mass distribution, initial spin, and slipping stage. — Reference 1
For a solid sphere, v_f = (5/7)v_i = 25/7 m/s; the source 1.2 m/s key is incorrect. — Reference 1

About this edition and its source ↗

Glossary and deeper reading · 7 refreshers

These editorial refreshers supplement the reading. Definitions and examples are added guidance; original source wording and recorded answers remain in their original context.

Mechanical work
Mechanical work is energy transferred by a force through displacement of its point of application. For a constant force on a particle, it depends on the force component along the particle’s displacement, and it can be positive, negative, or zero. For example, a magnetic force perpendicular to a particle’s displacement does zero work on that particle.
Read more: Work and force →
Kinetic energy
Kinetic energy is energy associated with motion relative to a chosen reference frame. For a nonrelativistic translating object it is one half of its mass times speed squared. For example, doubling a cart’s speed quadruples its translational kinetic energy when its mass stays fixed.
Read more: Kinetic energy →
Potential energy
Potential energy describes energy associated with the configuration of interacting parts of a system. Its numerical zero is a chosen reference, while differences in potential energy determine physical energy changes. For example, separating Earth and a ball changes their gravitational potential energy even though the energy belongs to their interaction.
Read more: Potential energy →
Gravitational potential energy
Gravitational potential energy is energy associated with the relative positions of attracting masses. Near Earth, changes can be approximated by mass times gravitational field strength times height change when that field is nearly constant. For example, lifting a book increases the gravitational potential energy of the book–Earth system.
Read more: Potential energy →
Rotational kinetic energy
Rotational kinetic energy is the kinetic energy associated with rotation about an axis. For a rigid body about a fixed axis it is one half of rotational inertia times angular speed squared. For example, a rolling wheel can possess both translational kinetic energy of its center of mass and rotational kinetic energy about that center.
Magnitude
Magnitude is the nonnegative size of a vector quantity. It does not specify direction, and it differs from a signed component. For example, a westward velocity can have a negative x component while its speed, the velocity magnitude, stays positive.
Vector
A vector is a quantity with a magnitude and a direction, represented by an arrow or by components along chosen axes. Vectors add by adding corresponding components rather than simply adding their magnitudes. For example, an eastward displacement followed by a northward displacement produces a resultant pointing northeast. The zero vector has zero magnitude and no unique direction.
Read more: Vector components and addition →

FFR - Rotational Energy

Rotational Kinetic Energy

Pop Quiz 6.5.1 Rotational Energy

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Chapter 6 Equations List

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Mechanical work

Kinetic energy

Potential energy

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Rotational kinetic energy

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