FOCUSED REVIEW
Focused Review — Rolling Motion — Algebra-Based
Reinforce the highest-leverage ideas and representative problem-solving tools for Rolling Motion.
TIME
Approximately 15 minutes
BEST FOR
Targeted reinforcement
FINISH WITH
A readiness check
After this focused review, you'll be able to...
reinforce the key relationships, apply them to representative problems, and identify what still needs work.
Choose how you want to review
Course Alignment
This Physics Sensei Unit Review is an independent learning resource. Use it to reinforce key concepts, prepare for homework, or review before a quiz or exam.
RESOURCE: Physics Sensei Unit Review | UNIT ID: MEC-U22 | TOPIC: Rolling Motion | COURSE LEVEL: Algebra-Based
BEST USED ✓ After learning the unit ✓ Before starting homework ✓ Before a quiz or exam
Your Review Plan
Complete these six stages in order. Each stage builds on the previous one and prepares you for the final readiness check.
6 Stages • Approximately 15 minutes.
Warm-Up Check
Before you begin, take a moment to see what you already remember. Do not worry about getting everything right. This is only a starting point.
ACTIVITY 1
Key Ideas
Write the no-slip relations for speed and tangential acceleration.
Write a short response: Write the no-slip relations for speed and tangential acceleration.
Reveal Answers
vCM = ωR and aCM = αR.
Why it works: vCM = ωR and aCM = αR.
ACTIVITY 2
Common Mistakes
What kinetic-energy term is often forgotten in rolling problems?
Write a short response: What kinetic-energy term is often forgotten in rolling problems?
Reveal Answers
The rotational term ½ Iω2.
Why it works: The rotational term ½ Iω2.
ACTIVITY 3
Quick Application
A wheel has R = 0.25 m and ω = 8.0 rad/s. Find vCM.
Write a short response: A wheel has R = 0.25 m and ω = 8.0 rad/s. Find vCM.
Reveal Answers
2.0 m/s.
Why it works: 2.0 m/s.
Ready to strengthen your understanding?
You've refreshed what you already know. Next, you'll reinforce the essential concepts that will help you solve problems with confidence. Need to see the learning path again?
Core Concepts
Reinforce the two highest-leverage relationships, then use them in representative situations.
KEY CONCEPT 1
No-Slip Kinematics
Pure rolling imposes vCM = ωR and aCM = αR. These equations let you convert between linear and angular variables.
vCM = ωR; aCM = αR
Example: If R = 0.30 m and ω = 10 rad/s, then vCM = 3.0 m/s.
Sensei note: Use the same radius that connects the center to the contact point.
KEY CONCEPT 2
Energy and Rolling Acceleration
With I = kMR2, energy and dynamics depend on k. For an object rolling down an incline without slipping, a = g sinθ /(1 + k).
I = kMR2; a = g sinθ/(1+k)
Example: Solid cylinder: k = ½, so a = (2/3)g sinθ.
Sensei note: Do not cancel rotational inertia; it is what distinguishes shapes.
KEY CONCEPT 3
Force and Torque Method
On an incline, use Mg sinθ - f = Ma and fR = Iα together with a = αR. With I = kMR2, these equations give a = g sinθ/(1+k) and the required static friction magnitude.
Mg sinθ − f = Ma; fR = Iα; a = αR
Example: For a solid cylinder, f = (1/3)Mg sinθ uphill and a = (2/3)g sinθ.
Sensei note: The sign of friction follows from your chosen axis and the tendency to slip; solve consistently rather than memorizing a direction.
Ready to apply these ideas?
You've reinforced the essential concepts. Now it's time to put them into practice by working through guided examples and building your problem-solving confidence. Need a quick reminder?
Guided Practice
Apply the reinforced ideas to two representative situations, then use the strategy card to check your setup.
PRACTICE 1
Guided Example
A solid sphere (k = 2/5) rolls down a 30° incline. Find a.
A solid sphere (k = 2/5) rolls down a 30° incline. Find a.
Reveal Answers
a = g sin30° /(1 + 2/5) = 4.9/1.4 = 3.5 m/s2.
Why it works: a = g sin30° /(1 + 2/5) = 4.9/1.4 = 3.5 m/s2.
PRACTICE 2
Independent Check
A hoop rolls from rest through a vertical drop h. Find its speed in terms of g and h.
A hoop rolls from rest through a vertical drop h. Find its speed in terms of g and h.
Reveal Answers
mgh = ½ Mv2 + ½(MR2)(v2/R2) = Mv2, so v = √(gh).
Why it works: mgh = ½ Mv2 + ½(MR2)(v2/R2) = Mv2, so v = √(gh).
PRACTICE 3
Independent Problem
A solid cylinder rolls without slipping down a 30° incline. Find a and the static-friction magnitude.
A solid cylinder rolls without slipping down a 30° incline. Find a and the static-friction magnitude.
Reveal Answers
a = (2/3)g sin30° = 3.27 m/s2. Using fR = Iα gives f = (1/3)Mg sin30° = Mg/6, directed uphill.
Why it works: a = (2/3)g sin30° = 3.27 m/s2. Using fR = Iα gives f = (1/3)Mg sin30° = Mg/6, directed uphill.
Ready to check your understanding?
You've practiced the essential skills with guidance. Now it's time to solve a few short problems on your own and confirm you're ready to move forward. Need a quick reminder?
Confidence Check
You've rebuilt the key ideas and practiced them with guidance. Now try these short questions on your own to check your understanding before moving on.
QUICK CHECK 1
Cylinder Speed
Answer without looking back.
A solid cylinder starts from rest and drops through height h. Write v at the bottom.
Reveal Answers
v = √(4gh/3).
Why it works: v = √(4gh/3).
QUICK CHECK 2
Incline Acceleration
Answer without looking back.
For I = kMR2, write the no-slip acceleration down an incline.
Reveal Answers
a = g sinθ /(1 + k).
Why it works: a = g sinθ /(1 + k).
QUICK CHECK 3
Friction Magnitude
Solve or explain without looking back.
For a solid cylinder on an incline, what fraction of Mg sinθ is the static-friction magnitude?
Reveal Answers
One third: f = (1/3)Mg sinθ.
Why it works: One third: f = (1/3)Mg sinθ.
How did it go?
You've checked your understanding. Take one final look at the essential ideas before deciding what to do next. Need a quick reminder?
Summary
Before moving on, take one final look at the most important ideas from this review.
KEY TAKEAWAY 1
Constraint
Use vCM = ωR and aCM = αR to connect translation and rotation.
KEY TAKEAWAY 2
Shape Factor
The factor k in I = kMR2 controls the energy split and acceleration.
KEY TAKEAWAY 3
Force and Torque Method
On an incline, use Mg sinθ - f = Ma and fR = Iα together with a = αR.
Ready for your next step?
You've reviewed the essential ideas one last time. Now choose the resource that best matches how confident you feel. Need a quick reminder?
Next Step
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