Problem 16 - College 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 of the shell and the friction force on it. Is the friction kinetic or static friction? Why? (b) How would your answers to part (a) change if the mass were doubled to 4.00 \(\mathrm{kg} ?\)

Step-by-Step Solution

Answer

The acceleration is 3.61 m/s², and the friction force is 3.61 N (static friction). Doubling the mass doesn't change these values.

Step 1: Identifying Forces and Parameters

The mass of the hollow spherical shell is given as \( m = 2.00 \, \text{kg} \) and the angle of the slope is \( \theta = 38.0^{\circ} \). Since the shell rolls without slipping, the frictional force here is static. We'll need to consider gravitational force, normal force, and friction.

Step 2: Using Newton's Second Law

The acceleration \( a \) of the center of mass of the sphere can be found using the equation: \[ a = \frac{g \cdot \sin(\theta)}{1 + I/(mR^2)} \] where \( I \) is the moment of inertia of the hollow sphere, \( I = \frac{2}{3}mR^2 \) (for a hollow sphere). Thus, \[ a = \frac{g \cdot \sin(\theta)}{1 + \frac{2}{3}} = \frac{g \cdot \sin(\theta)}{5/3} = \frac{3}{5} g \cdot \sin(\theta) \].

Step 3: Calculating Acceleration

Substitute \( g = 9.81\, \text{m/s}^2 \) and \( \theta = 38.0^{\circ} \): \[ \sin(38.0^{\circ}) \approx 0.615 \]. So, \[ a = \frac{3}{5} \times 9.81 \, \text{m/s}^2 \times 0.615 \approx 3.61 \, \text{m/s}^2 \].

Step 4: Calculating the Friction Force

The friction force can be determined by using the net torque and the rotational inertia. From the equation \( f = \frac{1}{2} m a \), we find \( f = \frac{1}{2} \times 2.00 \, \text{kg} \times 3.61 \, \text{m/s}^2 = 3.61 \, \text{N} \).

Step 5: Friction Type Justification

The friction is static because the sphere rolls without slipping, indicating there's no relative motion between the point of the sphere in contact with the slope and the slope itself.

Step 6: Analyzing Effects of Changing Mass

If the mass is doubled to 4.00 kg, the expressions for acceleration and friction force are independent of mass, assuming the same moment of inertia equation. Therefore, the acceleration and the friction force remain the same even if the mass is doubled.

Key Concepts

Static Friction

When an object rolls without slipping, like our hollow spherical shell on a slope, it relies on static friction. Static friction occurs when there is no relative motion between the object and the surface it's on. Unlike kinetic friction, which acts when surfaces slide against each other, static friction prevents slipping. This is crucial for rolling motion because it helps the object grip the surface.

In our problem, the hollow sphere rolls without slipping down a slope. Hence, the type of friction involved is static. This allows it to maintain continuous contact with the slope as it rolls smoothly. Static friction ensures that the bottom point of the shell, where it touches the slope, is momentarily stationary relative to the slope, keeping the rolling motion controlled and stable.

Remember that static friction adjusts itself to match the applied force up to a maximum limit, meaning it will increase as needed to prevent slipping until it reaches a threshold beyond which sliding would occur.

Newton's Laws of Motion

Newton's laws of motion are foundational for understanding the behavior of objects in motion. In our scenario of a rolling shell, they help explain how forces lead to the movement we observe.

According to Newton's second law of motion, the acceleration of an object depends on the net force acting upon it and its mass. This is captured in the formula:

• \( a = \frac{F_{net}}{m} \)

Here, the net force is influenced by gravitational force and static friction, and this helps us find the shell's acceleration as it rolls down the slope.
Newton's third law is also implicitly at play. As the shell pushes against the slope, the slope exerts an equal and opposite force back onto the shell, contributing to the rolling motion. These interactions follow the simple yet crucial principles of Newton's laws to describe the shell's movement seamlessly.

Moment of Inertia

The moment of inertia is a concept that helps us understand how an object resists changes in its rotational motion. For the hollow spherical shell, its moment of inertia affects how it accelerates as it rolls down the slope.

Moment of inertia depends on the object's mass and how this mass is distributed relative to the axis of rotation. For a hollow spherical shell, mathematically, the moment of inertia \( I \) is given by:

• \( I = \frac{2}{3}mR^2 \)

where \( m \) is the mass and \( R \) is the radius of the sphere.
In our exercise, the moment of inertia is integral in calculating the shell’s acceleration using the formula:

• \( a = \frac{g \cdot \sin(\theta)}{1 + I/(mR^2)} \)

As \( I \) determines how the mass is distributed, it ultimately decides how easily the shell rolls down the incline. The larger the moment of inertia, the more resistance there is to speeding up or slowing down the spin.

Gravitational Force

Gravitational force is the constant pull that the Earth exerts on objects, and it plays a significant role in rolling motion. For our rolling hollow spherical shell, gravity is the force that propels it down the slope.

Gravity acts vertically downwards, but when an object is on a slope, this force can be split into components. One component acts parallel to the slope and is given by \( mg \cdot \sin(\theta) \). This component is what helps accelerate the shell down the slope.

The other component acts perpendicular to the slope surface, providing the normal force but doesn't contribute directly to the shell’s movement along the slope. Understanding gravitational force and its components is vital to grasping how the shell gains speed as it descends, allowing us to correctly apply Newton’s laws to predict and explain its motion.