Problem 20
Question
A uniform solid sphere of mass \(m\) and radius \(r\) is placed on a ramp inclined at an angle \(\theta\) to the horizontal. The coefficient of static friction between sphere and ramp is \(\mu_{s} .\) Find the maximum value of \(\theta\) for which the sphere will roll without slipping, starting from rest, in terms of the other quantities.
Step-by-Step Solution
Verified Answer
Answer: θ_max = arctan(5 μs / (2 + 5 μs))
1Step 1: Draw a Free Body Diagram for the solid sphere
Draw a Free Body Diagram (FBD) of the sphere, illustrating the forces acting on it. The forces involved are gravitational force, static friction, and normal force. The gravitational force mg acts downward, normal force N acts perpendicular to the inclined plane, and the static friction fs acts up the inclined plane.
2Step 2: Newton's Second Law in the vertical direction
Apply Newton's second law in the vertical direction. The only force in this direction is the normal force. Therefore, we have:
N = mg * cos(theta)
3Step 3: Condition for rolling without slipping
To roll without slipping, the static friction force must be equal to the force required to rotate the sphere. Therefore, we have:
fs = m * a
where a is the acceleration of the sphere.
4Step 4: Force balance in the tangential direction
Apply Newton's second law in the tangential direction. Considering the forces acting on the sphere in this direction, we get:
m * a = mg * sin(theta) - fs
Substituting fs = m * a from step 3, we obtain:
m * a = mg * sin(theta) - m * a
5Step 5: Rotational dynamics
Consider the rotational dynamics of the sphere. The moment of inertia (I) for a solid sphere can be expressed as:
I = (2 / 5) * m * r^2
Using the angular acceleration \(\alpha\) and the relationship between linear and angular acceleration:
a = r * alpha
The torque acting on the sphere τ is defined by the relation:
τ = I * alpha
Substituting for I and α, we have:
fs * r = (2 / 5) * m * r^2 * (a / r)
6Step 6: Finding the maximum angle
Since the sphere is on the verge of slipping, we can equate the maximum value of static friction (fs_max = μs * N) to the expression for fs obtained in the previous step. Thus,
μs * N = (2 / 5) * m * r * a
Substitute for N:
μs * mg * cos(theta) = (2 / 5) * m * r * a
Now, substitute the expression for acceleration (a) from step 4:
μs * mg * cos(theta) = (2 / 5) * m * r * (mg * sin(theta) - fs)
which gives:
μs * cos(theta) = (2 / 5) * (sin(theta) - μs * cos(theta))
Now, solve for \(\theta\):
tan(theta) = (5 μs / (2 + 5 μs))
The maximum value for \(\theta\), denoted by \(\theta_{max}\), will be when tan(theta) has its highest value. Since we only need an expression in terms of the given quantities. The maximum value of \(\theta\) for which the sphere rolls without slipping starting from rest is achieved when:
θ_max = arctan(5 μs / (2 + 5 μs))
This is the answer to the exercise.
Key Concepts
Static FrictionInclined PlaneRotational DynamicsMoment of Inertia
Static Friction
Static friction is a force that needs to be overcome for an object to start moving from rest. This force exists because of the microscopic interactions between the two surfaces in contact; in our scenario, the solid sphere and the inclined plane.
In the rolling without slipping problem, static friction is what enables the sphere to roll up the inclined plane without sliding backward. It acts up the slope to oppose the component of gravitational force pulling the sphere down the slope. The stronger the static friction (characterized by the coefficient of static friction \( \mu_s \)), the steeper the incline can be for rolling to occur without slipping.
Static friction is limited and reaches its maximum value when it is equal to the product of the coefficient of static friction and the normal force (\( f_{s_{max}} = \mu_s \cdot N \)). It plays a critical role in determining the maximum angle \( \theta \) at which the sphere can start rolling without slipping.
In the rolling without slipping problem, static friction is what enables the sphere to roll up the inclined plane without sliding backward. It acts up the slope to oppose the component of gravitational force pulling the sphere down the slope. The stronger the static friction (characterized by the coefficient of static friction \( \mu_s \)), the steeper the incline can be for rolling to occur without slipping.
Static friction is limited and reaches its maximum value when it is equal to the product of the coefficient of static friction and the normal force (\( f_{s_{max}} = \mu_s \cdot N \)). It plays a critical role in determining the maximum angle \( \theta \) at which the sphere can start rolling without slipping.
Inclined Plane
An inclined plane, also known as a ramp, is a flat supporting surface tilted at an angle to the horizontal. This is a classic scenario used to study the components of forces and the dynamics of motion on a slope.
When an object is placed on an inclined plane, its weight can be split into two components - one perpendicular to the plane, which is balanced by the normal force, and one parallel to the plane, which is responsible for the object's tendency to slide down.
The angle of the incline (\( \theta \) in our exercise) directly affects these components. As the angle increases, the parallel component increases, making it more challenging for static friction to prevent slipping. It's important to find the balance between the force of gravity down the slope and the static friction to ensure rolling without slipping.
When an object is placed on an inclined plane, its weight can be split into two components - one perpendicular to the plane, which is balanced by the normal force, and one parallel to the plane, which is responsible for the object's tendency to slide down.
The angle of the incline (\( \theta \) in our exercise) directly affects these components. As the angle increases, the parallel component increases, making it more challenging for static friction to prevent slipping. It's important to find the balance between the force of gravity down the slope and the static friction to ensure rolling without slipping.
Rotational Dynamics
Rotational dynamics involves the study of objects in rotation, encompassing the relationships between torque, moment of inertia, and angular acceleration. In the context of our sphere rolling without slipping, we must consider how torque is needed to cause rotational acceleration.
Torque (\(\tau\)) is the rotational equivalent of force and is the product of the force causing the rotation (static friction in this case) and the distance from the pivot point to the line of action of the force (radius \(r\) of the sphere).
The equation \(\tau = I \cdot \alpha\) where \(I\) represents the moment of inertia, and \(\alpha\) is the angular acceleration, describes the rotational motion of the sphere. It is a crucial formula in solving for the angle of the incline as it links the linear and rotational aspects of motion.
Torque (\(\tau\)) is the rotational equivalent of force and is the product of the force causing the rotation (static friction in this case) and the distance from the pivot point to the line of action of the force (radius \(r\) of the sphere).
The equation \(\tau = I \cdot \alpha\) where \(I\) represents the moment of inertia, and \(\alpha\) is the angular acceleration, describes the rotational motion of the sphere. It is a crucial formula in solving for the angle of the incline as it links the linear and rotational aspects of motion.
Moment of Inertia
The moment of inertia (\(I\)) is a quantity expressing an object's tendency to resist angular acceleration; it is the rotational equivalent of mass in linear motion. The moment of inertia depends on the mass distribution relative to the axis of rotation. For instance, a solid sphere has a moment of inertia given by \(I = (2/5)mr^2\), where \(m\) is the mass and \(r\) is the radius.
In our physics problem, the moment of inertia plays a significant role in determining how easily the sphere can start to roll. A larger moment of inertia would mean that more torque is required to produce the same angular acceleration, affecting the static friction needed and consequently the maximum angle the incline can have for the sphere to roll without slipping.
In our physics problem, the moment of inertia plays a significant role in determining how easily the sphere can start to roll. A larger moment of inertia would mean that more torque is required to produce the same angular acceleration, affecting the static friction needed and consequently the maximum angle the incline can have for the sphere to roll without slipping.
Other exercises in this chapter
Problem 17
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