Unit 6 Notes: Rolling Motion and Orbital Motion (Energy + Angular Momentum Connections)

0.0(0)
Studied by 0 people
0%Unit 6 Mastery
0%Exam Mastery
Build your Mastery score
multiple choiceAP Practice
Supplemental Materials
call kaiCall Kai
Card Sorting

1/24

Last updated 3:12 PM on 3/12/26
Name
Mastery
Learn
Test
Matching
Spaced
Call with Kai

No analytics yet

Send a link to your students to track their progress

25 Terms

1
New cards

Rolling motion

Motion that combines translation of an object’s center of mass with rotation about its center of mass.

2
New cards

Translation (in rolling)

The forward motion of the center of mass (like a block sliding).

3
New cards

Rotation (in rolling)

Spinning motion about the object’s center of mass.

4
New cards

Rolling without slipping (pure rolling)

Rolling where the contact point with the surface is instantaneously at rest relative to the surface.

5
New cards

No-slip (rolling) constraint

For radius RR: vcm=θRv_{cm} = \theta R (and tangentially acm=θRa_{cm} = \theta R); links linear and rotational motion only when there is no slipping.

6
New cards

Static friction in pure rolling

Friction that prevents relative motion at the contact point; it can provide torque even though the contact point does not slide.

7
New cards

Why static friction often does no work in pure rolling

In the ground frame, the point of contact has zero displacement, so W = F·d at that point is zero (even though friction can provide torque).

8
New cards

Kinetic friction (slipping)

Friction that occurs when surfaces slide; it typically dissipates mechanical energy as thermal energy.

9
New cards

Total kinetic energy of a rolling object

Ktotal=12mvcm2+12Icmθ2K_{total} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}I_{cm}\theta^2 (translational plus rotational kinetic energy).

10
New cards

Moment of inertia about the center of mass (I_cm)

A measure of how mass is distributed relative to the rotation axis; determines how hard an object is to spin up.

11
New cards

Hoop (thin ring) moment of inertia

For radius RR about its symmetry axis: Icm=mR2I_{cm} = mR^2.

12
New cards

Solid disk/solid cylinder moment of inertia

For radius RR about its symmetry axis: Icm=12mR2I_{cm} = \frac{1}{2}mR^2.

13
New cards

Solid sphere moment of inertia

For radius RR about its symmetry axis: Icm=25mR2I_{cm} = \frac{2}{5}mR^2.

14
New cards

Thin spherical shell moment of inertia

For radius RR about its symmetry axis: Icm=23mR2I_{cm} = \frac{2}{3}mR^2.

15
New cards

Shape factor (β) for rolling

Defined by Icm=βmR2I_{cm} = \beta mR^2; larger β\beta means more rotational inertia relative to mR2mR^2.

16
New cards

Center-of-mass acceleration for rolling down an incline

For pure rolling: acm=g sinθ1+IcmmR2=g sinθ1+βa_{cm} = \frac{g \text{ sin} \theta}{1 + \frac{I_{cm}}{mR^2}} = \frac{g \text{ sin} \theta}{1+\beta}.

17
New cards

Instantaneous axis of rotation (rolling)

In pure rolling, the object can be treated as instantaneously rotating about the contact point (the axis changes continuously).

18
New cards

Speed of the top point of a rolling wheel

For pure rolling: vtop = vcm + ωR = 2v_cm (relative to the ground).

19
New cards

Orbit (physics definition)

Motion under gravity where an object continuously “falls around” a central body; gravity supplies the centripetal acceleration (it doesn’t disappear).

20
New cards

Circular orbital speed

For orbital radius rr around mass MM: v=GMrv = \sqrt\frac{GM}{r}; depends on rr and MM, not the satellite’s mass.

21
New cards

Centripetal acceleration in circular orbit

ac=v2r=GMr2a_c = \frac{v^2}{r} = \frac{GM}{r^2}; equals the local gravitational field magnitude for a circular orbit.

22
New cards

Orbital period for a circular orbit

T=2πrv=2πr3GMT = \frac{2\pi r}{v} = 2\pi\sqrt{\frac{r^3}{GM}}; larger rr gives a much longer period (Tr3/2T \propto r^{3/2}).

23
New cards

Total mechanical energy in a circular orbit

E = K + U = −GMm/(2r); negative for a bound orbit and becomes less negative (higher) as r increases.

24
New cards

Angular momentum conservation (satellites)

For a central force (gravity toward the center), torque about the center is zero, so angular momentum L = mr v_⊥ is conserved; implies faster speed when closer in an elliptical orbit.

25
New cards

Escape speed

Minimum speed at distance rr to reach infinity with zero final speed: ve=2GMr=2v_e = \frac{\sqrt{2GM}}{r} = \sqrt2 times the circular orbital speed at the same rr.