Wheel Rolling Uphill: Force Diagram & Angular Momentum

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A wheel with mass m and varying radii is rolling uphill on a slope with angle θ, influenced by a constant torque, Tau. To find the acceleration, the user plans to apply the equation sum(Tau) = I * alpha, using the moment of inertia I calculated for a solid disk, though they need to adjust it for the wheel's radial distribution. The discussion emphasizes the importance of incorporating the slope angle θ into the calculations. For angular momentum after one revolution, the user is reminded that L = I * ω and to use kinematic equations to relate angular acceleration and distance. Proper calculations and adjustments are necessary to solve both parts of the problem effectively.
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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.
 
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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?)
 
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