Problem 22 - University Physics with Modern Physics

Question

A hollow, spherical shell with mass 2.00 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 kg?

Step-by-Step Solution

Answer

Acceleration is 3.22 m/s², friction force is 6.44 N, \(\mu\) is 0.38. Doubling mass doubles friction force but doesn't affect \(a\) or \(\mu\).

Step 1: Understand the Problem

We need to find the acceleration, the friction force, and the minimum coefficient of friction for a hollow spherical shell rolling down a 38.0° slope without slipping. Then, we need to understand how these values change if the mass is doubled.

Step 2: Identify the Equations

For a hollow sphere rolling without slipping, we have the motion equation: \( a = \frac{g \sin(\theta)}{1 + \frac{I}{mr^2}} \)where \( I = \frac{2}{3}mr^2 \) for a hollow spherical shell. We also use Newton's second law and the rolling without slipping condition to find friction.

Step 3: Calculate the Moment of Inertia

The moment of inertia \( I \) of a hollow sphere is given by \( I = \frac{2}{3}mr^2 \). We keep it in terms of \( m \) and cancel where necessary in subsequent steps.

Step 4: Calculate the Acceleration

Substitute the hollow sphere's moment of inertia into the acceleration formula:\[ a = \frac{g \sin(38.0°)}{1 + \frac{2}{3}} = \frac{g \sin(38.0°)}{\frac{5}{3}} \approx \frac{3}{5} \times 9.81 \times \sin(38.0°) \approx 3.22 \, \text{m/s}^2 \]

Step 5: Find the Friction Force

Using Newton's second law for rotation and rolling without slipping, friction force \( f \) is\[ f = ma \]where we substitute for \( a \) calculated.\[ f = 2.00 \, \text{kg} \times 3.22 \, \text{m/s}^2 = 6.44 \, \text{N} \]

Step 6: Determine Minimum Coefficient of Friction

The minimum coefficient of friction \( \mu \) is\[ \mu mg \cos(38.0°) = f \]Solve for \( \mu \):\[ \mu = \frac{6.44}{2.00 \times 9.81 \times \cos(38.0°)} \approx 0.38 \]

Step 7: Evaluate Changes for Doubling the Mass

If the mass is doubled to 4.00 kg, the calculations for acceleration and coefficient of friction remain unchanged because these variables are independent of mass.However, the friction force becomes \[ f = 4.00 \, \text{kg} \times 3.22 \, \text{m/s}^2 = 12.88 \, \text{N} \]

Key Concepts

Moment of Inertia

In the context of rolling motion, the moment of inertia helps us understand how different shapes require different forces to rotate. The moment of inertia, often denoted by \( I \), essentially measures how much an object resists rotational acceleration.
This concept is similar to mass in linear motion. For a hollow spherical shell, the moment of inertia is given by the formula:

• \( I = \frac{2}{3}mr^2 \)

where \( m \) is the mass and \( r \) is the radius of the sphere.
The moment of inertia depends on the geometry of the object. A hollow sphere has a different moment of inertia compared to a solid sphere because the mass is distributed differently. This affects how the sphere rolls down the inclined plane, impacting the acceleration and friction forces involved.

Newton's Second Law

Newton's second law is crucial in understanding how forces affect the motion of objects. It states that the acceleration \( a \) of an object depends on the net force \( F \) acting upon it and its mass \( m \).
The formula is:

• \( F = ma \)

When applying Newton's second law to rolling motion, it accounts for both linear and rotational aspects. The acceleration of the center of mass of the sphere is determined by the component of gravitational force along the incline.
By considering rotational dynamics, another form of motion equation is used to find acceleration, considering both translational and rotational effects:

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

This equation highlights how the moment of inertia affects the rolling motion without slipping by modifying how the sphere accelerates as it goes down the slope.

Friction Force

Friction force plays a vital role in rolling motion, serving as the force that prevents slipping and allows the sphere to roll smoothly.
In this scenario, the friction force \( f \) is the force that resists the downward motion on the inclined plane. It's essential that this force is just enough to prevent slipping but not too much to cause skid.
Using Newton's second law and the calculated acceleration:

• \( f = ma \)

This force is crucial because it allows the sphere to roll without slipping by providing the necessary torque for rotation. In our example, the force was calculated to be 6.44 N for a sphere of 2.00 kg mass rolling down the slope.

Coefficient of Friction

The coefficient of friction \( \mu \) is a dimensionless number that describes the ratio of the force of friction between two bodies and the force pressing them together. In the case of rolling motion on an inclined plane, it determines how easily the sphere can roll down the slope.
The calculation is based on ensuring that the frictional force accommodates the required acceleration without causing slippage. It is derived from the balance of forces:

• \( \mu mg \cos \theta = f \)

Solving this gives the minimum coefficient of friction needed:

• \( \mu = \frac{f}{mg \cos \theta} \)

The calculation shows that a minimum coefficient of friction of approximately 0.38 is needed for the 2 kg sphere, ensuring it rolls without slipping.

Inclined Plane

An inclined plane is a flat surface tilted at an angle to the horizontal. It’s a simple machine that helps in changing the direction and magnitude of a force.
In physics problems, inclined planes are often used to analyze motion as they add a layer of complexity by incorporating gravitational components.Consider the angle of the incline, \( \theta = 38.0^\circ \), in our problem. This angle affects the gravitational component acting on the sphere:

• The parallel component is \( mg \sin \theta \), which causes motion down the incline.
• The perpendicular component is \( mg \cos \theta \), which relates to the normal force.

Working with inclined planes requires breaking down the gravitational force into these components, essential for solving dynamics problems involving rolling motion. This helps in understanding the interplay of forces causing the acceleration of the object while rolling.