AP Physics 1- 6.5 Rolling- Study Notes- New Syllabus
AP Physics 1-6.5 Rolling – Study Notes
AP Physics 1-6.5 Rolling – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
Rolling: Kinetic Energy of a System with Translational and Rotational Motion
Rolling Without Slipping
Rolling With Slipping
Rolling: Kinetic Energy of a System with Translational and Rotational Motion
A rolling object (like a wheel, ball, or cylinder) has both translational kinetic energy (due to the motion of its center of mass) and rotational kinetic energy (due to spinning about its axis).
Total Kinetic Energy:
For a rolling rigid body:
\( KE_{total} = KE_{translational} + KE_{rotational} \)
Translational KE:
\( KE_{trans} = \dfrac{1}{2} M v_{cm}^2 \)
Rotational KE:
\( KE_{rot} = \dfrac{1}{2} I \omega^2 \)
Thus, rolling motion combines both translation and rotation.
Special Case:
- If the shape and distribution of mass are known (moment of inertia depends on geometry), we can express the fraction of total KE in translation vs rotation.
Example:
For a solid sphere, \( I = \dfrac{2}{5}MR^2 \),
so: \( KE_{total} = \dfrac{7}{10} M v_{cm}^2 \)
Example :
A solid sphere of mass 2 kg and radius 0.2 m rolls without slipping at a speed of \( 5 \, \text{m/s} \). Calculate its total kinetic energy.
▶️Answer/Explanation
Step 1: Translational KE → \( \dfrac{1}{2} M v^2 = \dfrac{1}{2} (2)(25) = 25 \, \text{J} \).
Step 2: Rotational KE → \( \dfrac{1}{2} I \omega^2 \). For sphere, \( I = \dfrac{2}{5}MR^2 \), and \( \omega = \dfrac{v}{R} = \dfrac{5}{0.2} = 25 \, \text{rad/s} \).
\( KE_{rot} = \dfrac{1}{2} \cdot \dfrac{2}{5}(2)(0.2^2)(25^2) = 10 \, \text{J} \).
Step 3: Total KE = 25 + 10 = 35 J.
Example :
A hollow cylinder of mass 3 kg and radius 0.5 m rolls without slipping at \( 4 \, \text{m/s} \). Find its total kinetic energy.
▶️Answer/Explanation
Step 1: Translational KE → \( \dfrac{1}{2} (3)(4^2) = 24 \, \text{J} \).
Step 2: For hollow cylinder, \( I = MR^2 \), and \( \omega = v/R = 4/0.5 = 8 \, \text{rad/s} \).
\( KE_{rot} = \dfrac{1}{2} (3)(0.5^2)(8^2) = 24 \, \text{J} \).
Step 3: Total KE = 24 + 24 = 48 J.
Rolling Without Slipping
Rolling without slipping occurs when a rigid body (like a wheel, cylinder, or sphere) rolls on a surface such that the point of contact is momentarily at rest relative to the surface. In this case, the translational motion of the center of mass is directly related to the rotational motion of the system.
Main Points:
The translational velocity of the center of mass \(v_{cm}\) is related to the angular velocity \(\omega\) by:
\( v_{cm} = R \omega \) where \(R\) is the radius of the rolling object.
The acceleration of the center of mass \(a_{cm}\) is related to the angular acceleration \(\alpha\) by:
\( a_{cm} = R \alpha \)
- Friction is necessary to prevent slipping, but for ideal rolling without slipping, this friction does not dissipate energy (no work is done by static friction).
Rolling without slipping ensures that the total kinetic energy of the system is shared between translational and rotational motion:
\( KE_{total} = \dfrac{1}{2} M v_{cm}^2 + \dfrac{1}{2} I \omega^2 \)
- This condition is often assumed in problems to simplify calculations and avoid energy loss due to friction.
Example :
A solid cylinder of radius 0.3 m rolls without slipping at a speed of 6 m/s. Find its angular velocity.
▶️Answer/Explanation
Using rolling condition: \( v_{cm} = R \omega \)
\( \omega = \frac{v_{cm}}{R} = \frac{6}{0.3} = 20 \, \text{rad/s} \)
Answer: Angular velocity = 20 rad/s
Example :
A solid sphere rolls without slipping down a frictionless incline of height 2 m. Find the speed of its center of mass at the bottom.
▶️Answer/Explanation
Conservation of energy: Potential energy at top = Total kinetic energy at bottom
\( mgh = \dfrac{1}{2} M v_{cm}^2 + \dfrac{1}{2} I \omega^2 \)
For solid sphere, \( I = \dfrac{2}{5} MR^2 \) and \( \omega = v_{cm}/R \)
\( mgh = \dfrac{1}{2} M v_{cm}^2 + \dfrac{1}{2} \cdot \dfrac{2}{5} M v_{cm}^2 = \dfrac{7}{10} M v_{cm}^2 \)
\( v_{cm} = \sqrt{\dfrac{10}{7} gh} = \sqrt{\dfrac{10}{7} \cdot 9.8 \cdot 2} \approx 5.28 \, \text{m/s} \)
Rolling While Slipping
Rolling while slipping occurs when a rigid body (like a wheel, cylinder, or ball) rolls on a surface, but the point of contact slides relative to the surface. In this case, the translational and rotational motions are not directly related as in pure rolling, and kinetic friction acts to reduce slipping.
Main Points:
The translational velocity \(v_{cm}\) of the center of mass is not equal to the product of angular velocity and radius:
\( v_{cm} \neq R \omega \)
Sliding occurs at the point of contact, causing kinetic friction to act on the body. This friction:
- Opposes the relative motion at the contact point.
- Reduces the translational velocity of the center of mass or the rotational velocity depending on the situation.
- Dissipates mechanical energy as heat.
The motion can transition to rolling without slipping if friction slows the slipping enough to satisfy \( v_{cm} = R \omega \).
The total kinetic energy now includes losses due to friction:
\( KE_{total} = \dfrac{1}{2} M v_{cm}^2 + \dfrac{1}{2} I \omega^2 + \text{energy lost to friction} \)
Example :
A wheel is spinning at 20 rad/s while sliding forward on ice at 4 m/s. The radius of the wheel is 0.3 m. Determine whether it is rolling without slipping, and explain what happens over time.
▶️Answer/Explanation
For rolling without slipping: \( v_{cm} = R \omega = 0.3 \cdot 20 = 6 \, \text{m/s} \).
Since \( v_{cm} = 4 \neq 6 \, \text{m/s} \), the wheel is slipping.
Kinetic friction at the contact point will act to increase translational speed and decrease angular speed until \( v_{cm} = R \omega \), after which rolling without slipping occurs.
Example :
A solid cylinder is initially sliding with \(v_{cm} = 5 \, \text{m/s}\) and not spinning (\(\omega = 0\)) on a rough horizontal surface. The cylinder eventually rolls without slipping. Explain the motion.
▶️Answer/Explanation
Kinetic friction exerts a torque, causing the cylinder to start rotating. Simultaneously, friction reduces translational speed slightly.
Over time, \( v_{cm} \) and \( \omega \) adjust until the rolling condition \( v_{cm} = R \omega \) is satisfied.
After this, no slipping occurs, and friction no longer does work on the system.