6.5 Rolling

Rolling combines translational and rotational motion, making it a unique form of movement. Objects can roll without slipping, where their linear and angular motions are directly related, or with slipping, where these motions become decoupled.

Understanding rolling is crucial for analyzing real-world scenarios like wheels, balls, and cylinders in motion. It involves concepts of kinetic energy, friction, and the relationship between linear and angular quantities, providing a foundation for more advanced physics topics.

Kinetic energy of translational and rotational motion

Sum of translational and rotational kinetic energies

The total kinetic energy of a rolling object comes from both its linear motion and its rotation around its axis.

• Total kinetic energy equals the sum of translational and rotational energies: $K_{tot} = K_{trans} + K_{rot}$
• Translational kinetic energy depends on mass and velocity: $K_{trans} = \frac{1}{2}mv_{cm}^2$
• Rotational kinetic energy depends on moment of inertia and angular velocity: $K_{rot} = \frac{1}{2}I\omega^2$
• For a solid sphere rolling without slipping, this gives $K_{tot} = \frac{1}{2}mv_{cm}^2 + \frac{1}{2}(\frac{2}{5}mR^2)\omega^2$

Rolling without slipping

Relationship between translational and rotational motion

When an object rolls without slipping, its bottom point momentarily comes to rest against the surface as it rotates, creating a direct relationship between linear and rotational motion.

• Linear velocity relates to angular velocity: $v_{cm} = r\omega$
• Linear acceleration relates to angular acceleration: $a_{cm} = r\alpha$
• Linear displacement relates to angular displacement: $\Delta x_{cm} = r\Delta\theta$
• The instantaneous velocity at the point of contact with the ground is zero

Friction in ideal rolling

Static friction is essential for rolling without slipping, but interestingly it doesn't dissipate energy from the system.

• The contact point between object and surface is momentarily at rest
• Static friction provides the force needed to maintain rolling without slipping
• This friction acts perpendicular to the radius at the contact point, creating torque
• Unlike sliding friction, static friction doesn't reduce the system's mechanical energy

Rolling while slipping

Decoupled translational and rotational motion

When an object rolls while slipping, it slides along the surface while rotating, breaking the direct relationship between linear and rotational quantities.

• Linear velocity and angular velocity become independent: $v_{cm} \neq r\omega$
• The point of contact has non-zero velocity relative to the surface
• Linear and rotational motion must be analyzed separately using Newton's laws
• Eventually, kinetic friction will typically cause the object to transition to rolling without slipping

Energy dissipation due to friction

When an object slips while rolling, friction converts mechanical energy into thermal energy, gradually reducing the object's motion.

• Kinetic friction acts at the contact point, doing negative work on the system
• Energy is dissipated as heat, reducing total mechanical energy over time
• This energy loss affects both translational and rotational kinetic energies
• The system will eventually reach a state where $v_{cm} = r\omega$ (rolling without slipping)

Practice Problem 1: Kinetic Energy of Rolling Objects

Solution:

1. Identify the known quantities:

• Mass: $m = 2.0$ kg
• Radius: $r = 0.10$ m
• Linear velocity: $v_{cm} = 5.0$ m/s

1. Since the sphere is rolling without slipping, we can relate linear and angular velocities: $v_{cm} = r\omega$ $\omega = \frac{v_{cm}}{r} = \frac{5.0\text{ m/s}}{0.10\text{ m}} = 50\text{ rad/s}$

2. Calculate the translational kinetic energy: $K_{trans} = \frac{1}{2}mv_{cm}^2 = \frac{1}{2}(2.0\text{ kg})(5.0\text{ m/s})^2 = 25.0\text{ J}$

3. Calculate the rotational kinetic energy using the moment of inertia for a solid sphere: $I = \frac{2}{5}mr^2 = \frac{2}{5}(2.0\text{ kg})(0.10\text{ m})^2 = 0.008\text{ kg}\cdot\text{m}^2$ $K_{rot} = \frac{1}{2}I\omega^2 = \frac{1}{2}(0.008\text{ kg}\cdot\text{m}^2)(50\text{ rad/s})^2 = 10.0\text{ J}$

4. Calculate the total kinetic energy: $K_{tot} = K_{trans} + K_{rot} = 25.0\text{ J} + 10.0\text{ J} = 35.0\text{ J}$

Practice Problem 2: Rolling Down an Incline

Solution:

1. Identify the known quantities:

• Mass: $m = 5.0$ kg
• Radius: $r = 0.20$ m
• Initial velocity: $v_i = 0$ m/s
• Vertical displacement: $h = 2.0$ m

1. Apply conservation of energy. Initially, the cylinder has only gravitational potential energy. $E_i = mgh$

2. At the bottom, the energy is distributed between translational and rotational kinetic energy: $E_f = \frac{1}{2}mv_f^2 + \frac{1}{2}I\omega_f^2$

3. For a solid cylinder, the moment of inertia is: $I = \frac{1}{2}mr^2$

4. Since the cylinder rolls without slipping: $\omega_f = \frac{v_f}{r}$

5. Substitute into conservation of energy equation: $mgh = \frac{1}{2}mv_f^2 + \frac{1}{2}(\frac{1}{2}mr^2)(\frac{v_f}{r})^2$ $mgh = \frac{1}{2}mv_f^2 + \frac{1}{4}mv_f^2$ $mgh = \frac{3}{4}mv_f^2$ $v_f = \sqrt{\frac{4gh}{3}} = \sqrt{\frac{4(9.8\text{ m/s}^2)(2.0\text{ m})}{3}} = 5.1\text{ m/s}$

Practice Problem 3: Friction Force in Rolling

Solution:

1. Identify the known quantities:

• Mass: $m = 7.2$ kg
• Radius: $r = 0.108$ m
• Applied force: $F = 10$ N

1. Apply Newton's Second Law to find the linear acceleration: $F - f_s = ma_{cm}$

2. Apply the rotational version of Newton's Second Law: $\tau = I\alpha$ $f_s \cdot r = \frac{2}{5}mr^2 \cdot \alpha$

3. For rolling without slipping: $a_{cm} = r\alpha$

4. Solve for the friction force by combining these equations: $\alpha = \frac{a_{cm}}{r} = \frac{F - f_s}{mr}$ $f_s \cdot r = \frac{2}{5}mr^2 \cdot \frac{F - f_s}{mr}$ $f_s = \frac{2}{5}(F - f_s)$ $\frac{5}{5}f_s = \frac{2}{5}F$ $f_s = \frac{2}{7}F = \frac{2}{7}(10\text{ N}) = 2.86\text{ N}$