Problem 22 - University Physics with Modern Physics

Question

A hollow, spherical shell with mass 2.00 \(\mathrm{kg}\) rolls without slipping down a \(38.0^{\circ}\) slope. (a) Find the acceleration, the friction force, and the minimum coefficient of friction needed to prevent slipping. (b) How would your answers to part (a) change if the mass were doubled to 4.00 \(\mathrm{kg}\) ?

Step-by-Step Solution

Answer

(a) Acceleration = 4.55 m/s², friction force = 6.07 N, coefficient = 0.616. (b) Answers remain unchanged.

Step 1: Identify the forces

The forces acting on the spherical shell include the gravitational force, normal force, and frictional force. Set the gravitational force component along the slope (\( mg \sin \theta \)) and perpendicular to the slope (\( mg \cos \theta \)). Gravity provides the force to make it roll, while friction prevents slipping.

Step 2: Apply Newton's Second Law

Using Newton's second law along the slope, the net force causes the translational acceleration:\[ m a = m g \sin \theta - f \]where \( f \) is the frictional force. Using rotational motion, the torque due to friction about the center of the shell is \( \tau = f R = I \alpha \), where \( I = \frac{2}{3}mR^2 \) for a hollow sphere and \( \alpha = \frac{a}{R} \).

Step 3: Solve for acceleration

From the torque equation, we have:\[ f R = \frac{2}{3}m R^2 \frac{a}{R} \rightarrow f = \frac{2}{3}m a \]Substituting in Newton's second law, \[ m a = m g \sin \theta - \frac{2}{3}m a \]Solving for \( a \),\[ a = \frac{3}{5}g \sin \theta \]Plugging in the values:\[ a = \frac{3}{5} \times 9.8 \times \sin(38^{\circ}) = 4.55 \, \text{m/s}^2 \]

Step 4: Determine the friction force

Using the friction equation:\[ f = \frac{2}{3}m a \]Substitute the values,\[ f = \frac{2}{3} \times 2.00 \times 4.55 = 6.07 \, \text{N} \]

Step 5: Calculate the minimum coefficient of friction

For no slipping condition, the frictional force \( f = \mu mg \cos \theta \). Set\[ \mu mg \cos \theta = \frac{2}{3}mg \sin \theta \]Solving for \( \mu \):\[ \mu = \frac{2}{3} \frac{\sin \theta}{\cos \theta} = \frac{2}{3} \tan \theta \]\[ \mu = \frac{2}{3} \times \tan(38^{\circ}) = 0.616 \]

Step 6: Analyze effect of mass doubling

If the mass is doubled, the mass terms in the formulas for acceleration, friction force and minimum coefficient of friction would cancel out, resulting in the same values, meaning the acceleration, friction, and minimum coefficient of friction remain unchanged.

Key Concepts

Rolling Motion

Rolling motion is a combination of both translational and rotational movements. When an object, like our hollow spherical shell, rolls without slipping, it means that every point on the object in contact with the surface remains momentarily at rest relative to the surface before rotating.
This motion ensures a direct correlation between linear acceleration (\(a\)) and angular acceleration (\(\alpha\)).- A rolling object covers a distance along a surface just like it would if it were sliding.- Its rotational movement is related by the radius of the object, with condition \(v = \omega R\), where \(v\) is linear velocity, \(\omega\) is angular velocity, and \(R\) is the radius.- An example includes the spherical shell rolling down the slope, translating vertical motion into horizontal kinetic energy.The concept of rolling without slipping is critical in problems involving wheels, balls, and other circular objects on an incline.

Friction Force

Friction force keeps the object from slipping as it rolls down the incline. In this case, the friction is static and not kinetic because there’s no sliding between the spherical shell and the slope during motion. It acts at the point of contact and provides the necessary force for rolling without slipping:

• Opposes the direction of motion causing deceleration.
• Derived from the material's surface and the normal force acting on the object.
• In equilibrium with the component of gravitational force along the slope to prevent slipping.

Substituting the given conditions, the static friction ensures enough control so the shell rolls instead of slides.

Newton's Second Law

Newton's second law states that the net force acting on an object equals the mass of the object times its acceleration (\(F = ma\)).This fundamental principle applies both to linear and rotational dynamics in rolling motion scenarios:- Linearly, it controls the translational movement down the slope, balancing gravitational and frictional forces.
- For the shell, \(a\) (linear acceleration) results from redistribution of its weight along the slope and friction.
Utilizing Newton's laws, we derive linear acceleration as\[ a = \frac{3}{5}g \sin \theta \]This relationship helps in calculating how quickly the object gathers speed as it rolls down the slope.

Torque

Torque is the measure of the force that can cause an object to rotate about an axis. In this exercise, torque is produced by the friction force acting on the shell, which influences its angular acceleration:- Torque (\(\tau\)) equals the friction force times the radius of the object (\(\tau = fR\)).- This sets the rotational motion according to the shell's moment of inertia (\(I\)) and relationship \(\tau = I\alpha\).

• For a hollow sphere, \(I = \frac{2}{3}mR^2\).
• \(\alpha = \frac{a}{R}\).
• Balancing these, \(f \times R = I \times \alpha\).

The concept of torque assists in understanding how the frictional force affects angular motion directly, linking linear and rotational dynamics.

Coefficient of Friction

The coefficient of friction (\(\mu\)) is a dimensionless value that represents the frictional force magnitude between two surfaces in contact.In the exercise, it determines how much friction is necessary to prevent slipping:- Static friction proportionally relates to this coefficient with the normal force as
\[ f = \mu mg \cos \theta \]- Allows us to understand the minimum \(\mu\) to maintain static friction using\[ \mu = \frac{2}{3} \tan \theta \]Calculating and knowing such coefficients help in designing surfaces and materials to optimize rolling efficiency and minimize unnecessary slips.