Wheel Rolling Uphill: Force Diagram & Angular Momentum

[laurids]

Homework Statement

A wheel with mass m, inner radius r and outer radius R is placed on a slope. The slope has angle θ related to horizontal. When the wheel is affected by a constant, known torque Tau in the centre of mass, the wheel begins rolling uphill.

http://sveskekat.dk/files/uploads/phys3.PNG

a) I need to draw a force diagram (free body diagram?) and find the acceleration of the wheel

b) The wheel starts at rest and rolls uphill. What is the angular momentum when the wheel has rolled one revolution?

Homework Equations

sum(Tau) = I * alpha

The Attempt at a Solution

I did not attempt b) yet, so please disregard that one so far.

For a) I am thinking I can use sum(Tau) = I * alpha.
I did a calculation for I of the wheel, but doing so I assumed that the wheel has even density, and is a solid disk. I don't know if it is correct to assume this, but I tried it, and I then replaced sum(Tau) for Tau and alpha for alpha = a / R, and then solve for a.

But I think I need to use the angle θ of the slope somehow. I need suggestions.
Thanks.

 

[LowlyPion]

Welcome to PF.

Your idea is right with respect to using τ to determine the angular acceleration α.

τ = I*α

You will need however to calculate a proper I based on the given that the wheel, while evenly distributed, is only distributed over the radial range of r to R.

∫ ρ*2πr*r² dr from r to R

(Solving the integral requires recognizing the total mass m is given by the mass area density ρ times the area of the ring which is π(R² - r²). )

For b) you know that

L = I*ω

and you should know from linear kinematics analogy that

Final-ω² = Initial-ω² + 2*α * x

where α is your angular acceleration and x is the distance in radians. (1 revolution is how many radians again?)

 

[nlightner]

I would approach this problem by first drawing a force diagram or free body diagram of the wheel on the slope. This will help to visually represent all the forces acting on the wheel and determine their magnitudes and directions.

From the diagram, we can see that there are two main forces acting on the wheel: the weight of the wheel (mg) and the normal force from the slope (N). The weight acts downwards and the normal force acts perpendicular to the slope. Additionally, there is a torque (Tau) acting on the wheel at the center of mass, causing it to roll uphill.

Using the equation sum(Tau) = I * alpha, we can solve for the acceleration (a) of the wheel. However, we also need to take into account the angle of the slope (θ) in our calculation. We can do this by using the component of the weight and normal force that act parallel to the slope. This can be represented by mg*sin(θ) and N*cos(θ).

Our equation then becomes sum(Tau) = (I + mR^2)*alpha = (mg*sin(θ) - N*cos(θ)) * R.

Solving for alpha, we get alpha = (mg*sin(θ) - N*cos(θ)) * R / (I + mR^2).

This will give us the angular acceleration of the wheel. To find the linear acceleration (a), we can use the equation a = alpha * R, where R is the radius of the wheel.

For part b), we can use the equation for angular momentum, L = I * omega, where omega is the angular velocity of the wheel. Since the wheel starts at rest and rolls one revolution, we can use the equation L = mR^2 * omega = mR^2 * 2π / T, where T is the time it takes for the wheel to make one revolution. This will give us the angular momentum of the wheel after one revolution.

Overall, the key to solving this problem is to carefully consider all the forces and their components acting on the wheel and incorporating the angle of the slope in our calculations.