AP Physics C Mechanics- 6.5 Rolling- Study Notes- New Syllabus
AP Physics C Mechanics- 6.5 Rolling – Study Notes
AP Physics C Mechanics- 6.5 Rolling – Study Notes – per latest Syllabus.
Key Concepts:
- Kinetic Energy of a System with Translational and Rotational Motion
- Rolling Without Slipping
- Slipping and Energy Dissipation in Rolling Motion
Kinetic Energy of a System with Translational and Rotational Motion
A rigid body in motion can have two types of kinetic energy:
- Translational kinetic energy — due to motion of the center of mass through space.
- Rotational kinetic energy — due to rotation about an axis (often through its center of mass).
The total kinetic energy of such a system is the sum of these two forms:
\( \mathrm{K_{tot} = K_{trans} + K_{rot}} \)
or equivalently,
\( \mathrm{K_{tot} = \tfrac{1}{2} M v_{cm}^2 + \tfrac{1}{2} I_{cm} \omega^2} \)
- \( \mathrm{M} \): total mass of the system
- \( \mathrm{v_{cm}} \): speed of the center of mass
- \( \mathrm{I_{cm}} \): moment of inertia about the center of mass
- \( \mathrm{\omega} \): angular velocity about the rotational axis
Key Idea: A rolling or rotating object has both linear motion (movement of its center of mass) and rotational motion (spinning about its axis). The total energy accounts for both components of motion simultaneously.
Relationship Between Translational and Rotational Motion (Rolling Without Slipping)
For a rolling object where there is no slipping, the translational velocity of the center of mass and the angular velocity are related by:
\( \mathrm{v_{cm} = R\omega} \)
Substituting into the total kinetic energy expression:
\( \mathrm{K_{tot} = \tfrac{1}{2} M v_{cm}^2 + \tfrac{1}{2} I_{cm} \dfrac{v_{cm}^2}{R^2}} \)
This shows how the distribution of mass (through \( \mathrm{I_{cm}} \)) affects the total energy of a rolling object.
Example
A solid cylinder of mass \( \mathrm{2.0\,kg} \) and radius \( \mathrm{0.20\,m} \) rolls without slipping along a horizontal surface at a speed of \( \mathrm{3.0\,m/s} \). Find its total kinetic energy.
▶️ Answer / Explanation
Step 1: Identify components of kinetic energy:
\( \mathrm{K_{tot} = K_{trans} + K_{rot}} \)
Step 2: Compute each term:
\( \mathrm{K_{trans} = \tfrac{1}{2} M v^2 = \tfrac{1}{2}(2.0)(3.0)^2 = 9.0\,J} \)
Moment of inertia for a solid cylinder: \( \mathrm{I_{cm} = \tfrac{1}{2}MR^2} \)
\( \mathrm{K_{rot} = \tfrac{1}{2} I_{cm}\omega^2 = \tfrac{1}{2}(\tfrac{1}{2}MR^2)(\tfrac{v^2}{R^2}) = \tfrac{1}{4}Mv^2} \)
\( \mathrm{K_{rot} = \tfrac{1}{4}(2.0)(3.0)^2 = 4.5\,J} \)
Step 3: Add both parts:
\( \mathrm{K_{tot} = 9.0 + 4.5 = 13.5\,J} \)
Result: The total kinetic energy of the rolling cylinder is \( \mathrm{13.5\,J} \).
Rolling Without Slipping
When a rigid body rolls without slipping on a surface, there is a fixed relationship between the linear motion of the center of mass and the rotational motion of the object about its axis.
For a rolling object, the point of contact with the surface is instantaneously at rest relative to the surface — this condition defines rolling without slipping.
Key Relationships:
\( \mathrm{\Delta x_{cm} = R\,\Delta \theta} \)
\( \mathrm{v_{cm} = R\,\omega} \)
\( \mathrm{a_{cm} = R\,\alpha} \)
- \( \mathrm{R} \): radius of the rolling object
- \( \mathrm{\Delta x_{cm}} \): displacement of the center of mass
- \( \mathrm{\omega} \): angular velocity
- \( \mathrm{\alpha} \): angular acceleration
Key Idea: The linear displacement, velocity, and acceleration of the object’s center of mass are directly proportional to the corresponding angular quantities through the radius \( \mathrm{R} \). This connection ensures that the object rolls smoothly without slipping or sliding.
Visual Interpretation:
- The point on the rim touching the ground has zero velocity instantaneously.
- The center of mass moves forward at speed \( \mathrm{v_{cm}} \).
- The top of the object moves at speed \( \mathrm{2v_{cm}} \) relative to the ground.
Example
A solid sphere of radius \( \mathrm{0.10\,m} \) rolls without slipping along a flat surface. If its angular speed is \( \mathrm{20\,rad/s} \), find the translational speed of its center of mass and the distance it travels in \( \mathrm{2.0\,s} \).
▶️ Answer / Explanation
Step 1: For rolling without slipping, \( \mathrm{v_{cm} = R\omega} \).
\( \mathrm{v_{cm} = (0.10)(20) = 2.0\,m/s} \)
Step 2: The distance traveled in \( \mathrm{2.0\,s} \):
\( \mathrm{\Delta x_{cm} = v_{cm}t = (2.0)(2.0) = 4.0\,m} \)
Result: The center of mass moves at \( \mathrm{2.0\,m/s} \) and travels \( \mathrm{4.0\,m} \) in \( \mathrm{2.0\,s} \).
Energy in Rolling Without Slipping (Ideal Case)
In ideal rolling without slipping, the static frictional force at the point of contact prevents slipping, but it does no work on the rolling object.
This is because the point of contact is instantaneously at rest with respect to the surface there is no relative motion at that point, and thus no energy dissipation due to friction.
Key Idea:
- The frictional force is static, not kinetic.
- It provides the necessary torque for rolling, but does not convert mechanical energy into heat.
- Total mechanical energy (kinetic + potential) remains conserved if no other nonconservative forces act.
Energy Expression (for ideal rolling):
\( \mathrm{K_{total} = K_{trans} + K_{rot} = \tfrac{1}{2} M v_{cm}^2 + \tfrac{1}{2} I_{cm} \omega^2} \)
Since \( \mathrm{v_{cm} = R\omega} \), both forms of motion are linked without any loss of energy to friction.
Example
A solid cylinder rolls without slipping down a 2.0 m high incline. Assuming no energy loss due to friction, find the cylinder’s translational speed at the bottom.
▶️ Answer / Explanation
Step 1: Apply energy conservation:
\( \mathrm{Mgh = \tfrac{1}{2} M v_{cm}^2 + \tfrac{1}{2} I_{cm} \omega^2} \)
For a solid cylinder, \( \mathrm{I_{cm} = \tfrac{1}{2}MR^2} \) and \( \mathrm{v_{cm} = R\omega} \).
Step 2: Substitute and simplify:
\( \mathrm{Mgh = \tfrac{1}{2}Mv_{cm}^2 + \tfrac{1}{2}(\tfrac{1}{2}MR^2)(\tfrac{v_{cm}^2}{R^2})} \)
\( \mathrm{Mgh = \tfrac{3}{4}Mv_{cm}^2} \)
Step 3: Solve for \( \mathrm{v_{cm}} \):
\( \mathrm{v_{cm} = \sqrt{\dfrac{4gh}{3}} = \sqrt{\dfrac{4(9.8)(2.0)}{3}} = 5.1\,m/s} \)
Result: The cylinder reaches the bottom with \( \mathrm{v_{cm} = 5.1\,m/s} \), with energy distributed between translational and rotational motion but no loss due to friction.
Motion of a System’s Center of Mass During Slipping
When a rotating system slips instead of rolling smoothly, the translational motion of the system’s center of mass and its rotational motion are no longer directly related.
In rolling without slipping, the condition \( \mathrm{v_{cm} = R\omega} \) holds true because the point of contact is instantaneously at rest. However, when slipping occurs, this condition breaks down:
\( \mathrm{v_{cm} \neq R\omega} \)
- If \( \mathrm{v_{cm} > R\omega} \), the object is sliding forward (spinning too slowly).
- If \( \mathrm{v_{cm} < R\omega} \), the object is skidding backward (spinning too fast).
Key Idea: Because of slipping, there is relative motion between the object and the surface at the contact point. This relative motion introduces kinetic friction, which affects both linear and rotational motion differently.
Effect on Motion:
- The center of mass may accelerate or decelerate depending on the direction of kinetic friction.
- Angular velocity changes due to the torque exerted by kinetic friction.
- Unlike rolling, the linear and angular accelerations cannot be related by \( \mathrm{a_{cm} = R\alpha} \).
Example
A bowling ball is thrown so that initially it slides without rolling on the lane. Its initial linear speed is \( \mathrm{8.0\,m/s} \), and its angular speed is \( \mathrm{0\,rad/s} \). The coefficient of kinetic friction between the ball and the lane is \( \mathrm{0.20} \). Explain how the motion of the ball changes before it begins to roll without slipping.
▶️ Answer / Explanation
Step 1: Since the ball is slipping, \( \mathrm{v_{cm} \neq R\omega} \). Initially, \( \mathrm{v_{cm} > R\omega} \) because it slides without rotation.
Step 2: Kinetic friction acts backward on the ball’s translational motion, reducing \( \mathrm{v_{cm}} \), and simultaneously exerts a torque that increases \( \mathrm{\omega} \).
Step 3: Eventually, as friction continues to act:
\( \mathrm{v_{cm}} \downarrow \text{ and } \omega \uparrow \)
until the condition \( \mathrm{v_{cm} = R\omega} \) is satisfied.
Step 4: At that instant, the ball transitions to pure rolling without slipping.
Result: Kinetic friction reduces the translational speed and increases rotational speed until rolling without slipping begins.
Energy Dissipation Due to Kinetic Friction During Slipping
When a rotating system slips relative to a surface, kinetic friction acts at the point of contact. Unlike static friction, which does no work during pure rolling, kinetic friction does work because the contact point moves relative to the surface.
Therefore, kinetic friction dissipates mechanical energy from the system as thermal energy (heat):
\( \mathrm{W_{fric} = -f_k \Delta x_{contact}} \)
where
- \( \mathrm{f_k = \mu_k N} \): kinetic frictional force
- \( \mathrm{\Delta x_{contact}} \): displacement of the point of contact relative to the surface
Note:
- The mechanical energy (kinetic + potential) of the rolling system decreases.
- The lost energy appears as heat due to frictional work.
- The direction of kinetic friction always opposes relative motion between the surfaces.
Example
A \( \mathrm{2.0\,kg} \) solid sphere (radius \( \mathrm{0.10\,m} \)) is sliding on a horizontal surface with \( \mathrm{v_{cm} = 6.0\,m/s} \) and \( \mathrm{\omega = 0} \). The coefficient of kinetic friction is \( \mathrm{0.30} \). Determine how much mechanical energy is lost when the sphere transitions to rolling without slipping.
▶️ Answer / Explanation
Step 1: For pure rolling, the condition is \( \mathrm{v_{f} = R\omega_f} \).
For a solid sphere: \( \mathrm{I = \tfrac{2}{5}MR^2} \)
Step 2: Using energy and angular momentum considerations, the fraction of translational kinetic energy lost is known to be:
\( \mathrm{\dfrac{\Delta K}{K_i} = \dfrac{1}{7}} \)
Step 3: Initial kinetic energy:
\( \mathrm{K_i = \tfrac{1}{2} M v_i^2 = \tfrac{1}{2}(2.0)(6.0)^2 = 36.0\,J} \)
Step 4: Energy lost to friction:
\( \mathrm{\Delta K = \tfrac{1}{7}K_i = \tfrac{1}{7}(36.0) = 5.14\,J} \)
Result: Approximately \( \mathrm{5.1\,J} \) of mechanical energy is converted into heat due to kinetic friction as the sphere transitions to rolling without slipping.