Rotational Equilibrium and Rotational Dynamics problem

In summary, a potter's wheel with a radius of 0.50 m and a moment of inertia of 12 kg m^2 is rotating at 50 rev/min. By pressing a wet rag against the rim with a radially inward force of 70 N, the potter can stop the wheel in 6.0 sec. To find the effective coefficient of kinetic friction between the wheel and the rag, the angular version of Newton's 2nd law must be used. The friction force, which is the tangential force producing the torque that stops the wheel, is related to the normal force exerted by the potter.
  • #1
BBallman_08
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The Question:

A potter's wheel having a radius of 0.50 m and a moment of inertia of 12 kg m^2 is rotating freely at 50 rev/min. The potter can stop the wheel in 6.0 sec by pressing a wet rag against the rim and exerting a radially inward force of 70 N. Find the effective coefficient of kinetic friction between the wheel and the rag.

What I know:

I know with the info given, I must use the angular version of Newton's 2nd law.

Sum (Torque) = I(Moment of Inertia) Alpha (Angular Acceleration).

But I am unsure how this even relates to the coeff. of kinetic friction.

Any help?
Thanks In Advance
 
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  • #2
The friction force is the tangential force producing the torque that stops the wheel. How does kinetic friction relate to the normal force, which is the force with which the potter pushes against the rim?
 
  • #3




Great job using the angular version of Newton's 2nd law! This problem involves rotational equilibrium and rotational dynamics, which means we need to consider both the forces and the moments acting on the system. In this case, the potter's wheel is rotating freely at a constant speed, indicating that it is in rotational equilibrium. However, when the potter applies a force to the wheel with the wet rag, the wheel experiences a change in its angular velocity, indicating that rotational dynamics are at play.

To find the coefficient of kinetic friction between the wheel and the rag, we can use the equation for torque: τ = rFsinθ, where τ is the torque, r is the radius of the wheel, F is the force applied, and θ is the angle between the force and the radius. In this case, the force applied is the 70 N force exerted by the potter, and the angle θ is 90 degrees (since the force is applied radially inward). We can rearrange this equation to solve for the coefficient of kinetic friction:

μ = τ/(rFsinθ)

Substituting in the given values, we get:

μ = (12 kg m^2 * α)/(0.50 m * 70 N * sin 90°)

Next, we need to find the angular acceleration α of the wheel. We can use the equation α = Δω/Δt, where ω is the final angular velocity and Δt is the time it takes for the wheel to stop (6.0 seconds). We know that the initial angular velocity is 50 rev/min, which we need to convert to radians per second (ω = (50 rev/min)*(2π rad/rev)*(1 min/60 s) = 5.24 rad/s). Plugging in these values, we get:

μ = (12 kg m^2 * 5.24 rad/s)/(0.50 m * 70 N * 1)

Solving for μ, we get a value of approximately 0.018. This is the effective coefficient of kinetic friction between the wheel and the rag. Keep in mind that this value may vary depending on the specific materials and conditions of the wheel and rag. I hope this helps!
 

1. What is rotational equilibrium?

Rotational equilibrium refers to a state in which an object is not rotating or spinning, but instead remains at rest or maintains a constant rotational speed.

2. How is rotational equilibrium different from translational equilibrium?

Translational equilibrium refers to a state in which an object is not moving or is moving at a constant velocity, while rotational equilibrium refers to a state in which an object is not rotating or is rotating at a constant speed. In translational equilibrium, all forces acting on the object must be balanced, while in rotational equilibrium, both the forces and torques (rotational forces) must be balanced for an object to remain at rest or maintain a constant rotational speed.

3. What factors affect the rotational equilibrium of an object?

The rotational equilibrium of an object is affected by the object's mass, shape, and distribution of mass, as well as the forces acting on the object and the distance at which those forces are applied. Additionally, the object's moment of inertia, which is a measure of its resistance to rotational motion, plays a significant role in determining its rotational equilibrium.

4. What is the difference between rotational dynamics and rotational equilibrium?

Rotational dynamics refers to the study of how forces and torques act on an object to produce rotational motion, while rotational equilibrium refers to a specific state in which an object is not experiencing any net torque and therefore remains at rest or maintains a constant rotational speed. Rotational equilibrium is a subset of rotational dynamics, as it only applies to a specific state of motion.

5. How can I calculate the torque required to maintain rotational equilibrium?

The torque required to maintain rotational equilibrium can be calculated using the formula τ = rFsin(θ), where τ is the torque, r is the distance from the axis of rotation to the point where the force is applied, F is the force, and θ is the angle between the force vector and the lever arm. This formula is based on the principle of moments, which states that the sum of all torques acting on an object in rotational equilibrium must be equal to zero.

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