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Two cylindrical jumping wheels

1 vote
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A cart has two cylindrical wheels connected by a
weightless horizontal rod using weightless spokes and friction-
less axis as shown in the figure. Each of the wheels is made
of a homogeneous disc of radius R, and has a cylidrical hole
of radius R/2 drilled coaxially at the distance R/3 from the
centre of the wheel. The wheels are turned so that the holes
point towards each other, and the cart is put into motion on
a horizontal floor. What is the critical speed v by which the
wheels start jumping?

My attempt:
I have noticed that while rolling at constant speed, the centre of
mass of the whole cart moves also with a constant speed, i.e.
there should be no horizontal forces acting on the cart. Also,
each of the cylinders rotates with a constant angular speed,
hence there should be no torque acting on it, hence the friction force must be zero.

My book also gave a hint which I couldn't interpret-Use the rotating frame of a wheel;
Try to substitute one asymmetric body
(the cylinder with a hole) with two symmetric bodies, a hole-
less cylinder, and a superimposed cylinder of negative density.
Keep in mind that the rod can provide any horizontal
force, but cannot exert any vertical force.

The answer = $$v = {3\sqrt(g)\sqrt(R)}$$

asked May 29, 2017 in Physics Problems by Takeover (130 points)
edited May 30, 2017 by Takeover
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1 Answer

1 vote

If the cart is rolling by itself, its mechanical energy (KE+PE) will be constant but its linear and angular velocities will not be constant. Its motion will be jerky because the CM moves up and down so the KE is not constant.

I think we must assume that the cart is pulled along at constant velocity. The pulling force will vary, to ensure that the cart moves at constant velocity. However, we only need to consider vertical forces to find the speed at which the wheels "jump" upwards. We must assume that the pulling force has no vertical component, which could affect "jumping".

The CM of the wheels rotates in a circle of radius $r$ about the geometrical centre of the wheel (the centroid). You can find $r$ by considering the centre of mass of the solid disk and the part which is removed.

As the cart rotates, centripetal force of $Mr\omega^2$ is required to keep the CM moving in a circle, where $M$ is the mass of the wheels and $v=R\omega$ is the speed of the cart. At the top of this circle the only force directed towards the centroid is gravity. If gravity is not sufficient then the wheels will "jump" off the ground. So we must have :
$Mr\omega^2 \le Mg$.
From this you can work out the maximum speed of the cart to avoid the wheels "jumping".

answered May 30, 2017 by sammy gerbil (28,746 points)
edited May 30, 2017 by sammy gerbil
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