Rolling wheel question

Question

A wheel made up of a rim (a ring having mass m and radius R)

along with eight uniform rods, each of mass m but not all of the same length).

If the wheel rolls without slipping down a fixed incline that makes an angle= 10 with

the ground, determine the acceleration of the center of the wheel down the incline.

Answer 1

The situation is as follows : An object with a circular cross section (in this case, a wheel with a rim of mass M and 8 spokes also each of M) of mass 9 M and radius R rolls without slip down an incline of angle θ. We are interested in finding the acceleration a of the wheel as it rolls down the incline.

I have already written in detail about the physics of rolling without slipping under the influence of a force as well as static friction on another question (link in comment), so you can check out the answer to that question to understand how to conceptualize this problem.

First, let us list out the forces acting on the wheel :

1) Force due to gravity 9 Mg acting vertically down at the center of mass (C.O.M). This may be resolved into two components : 9 Mg sin θ down the incline, ang 9 Mg cos θ ⊥ into the surface of the incline.

2) Normal force N exerted by the surface of the incline. N = 9 Mg cos θ along outward ⊥ of the surface acting at .P_touch , the point at which at which the wheel touches the surface)

3) Static frictional force f_s acting up the incline at P_touch (as in rolling without slipping, the point P_touch at which the wheel touches the surface at any given instant is (momentarily) at rest at that very instant. The state of rest is maintained by the static friction between the surface of the incline and the rolling surface of the wheel)

The forces along the surface of the incline result in acceleration of the C.O.M down the incline with acceleration a, so we may write :

9 Mg sin θ - f_s = 9 Ma

Now, let us analyze the rotational component of the motion. Rolling without slipping means that as the C.O.M is accelerating with a, the wheel also experiences angular acceleration at rate α = a / R. So now, the task is to evaluate the torque τ_P about some convenient point P on the wheel, and relate that to α using the angular version of Newton's 2nd law , τ_P = I_P α , where I_P is the moment of inertia of the wheel about the point P.

Here, the convenient choice of P = C.O.M. About the C.O.M, one can see (by drawing a diagram) that only the frictional force f_s exerts a torque. So, τ_C.O.M = f_s R. So the main challenge is to find the moment of inertia I_C.O.M

I_C.O.M = I_C.O.M^rim + 8 I_C.O.M^spoke, where I_C.O.M^rim is the contribution of the rim to the total moment of inertia of the wheel about the C.O.M, and I_C.O.M^spoke is the contribution of 1 spoke. I_C.O.M^rim = MR², because the rim is an object which has all its mass concentrated at a distance R from the rotating axis under consideration (the C.O.M).

For objects which have their masses distributed over a range of distances from the rotating axis (solid objects), upto a maximum of R, the I = k MR², where k < 1 is a factor that indicates how closely to the rotating axis the mass of the object is concentrated. To find k, we should break up this extended object into infinitesimal masses dm (which are so small that we can think of them as being 'points' with negligible spatial extent), and sum the contributions dm r² from each infinitesimal piece, where r is the distance of the location of the mass element dm from the rotating axis.

In other words, the moment of inertia I is the volume integral I = ∫ dm r² ('Volume integral' because dm = ρ dV, where dV is an infinitesimal volume element and ρ is the density).

In the case of the rod attached to the C.O.M on one end, volume elements of length dr weigh dm = M / R dr (as the mass per unit length of the rod of mass M and length R is M / R).

So, I_C.O.M^spoke = (M / R) ∫₀^R dr r² = (1 / 3) MR².

Putting everything together, you get 2 equations, which may be used to evaluate the unknowns (f_s and a) in terms of the knowns in this problem (M , R , θ)

Hope this helps.