Intriguing problem from Rotational Mechanics ?

Question

A child gets hold of a thin ring of inner radius '3πR', outer radius '4πR' and thickness 'R/(2π²)'. The mass of the disc is 'M' and it has a uniform mass distribution.

The child thinks of balancing the ring on the ground. The coefficient of friction between the ground and the ring is 'μ'. At first he simply places the ring on the ground but obviously, as soon as he does this the ring topples over. So, in order to keep the ring balanced on the ground for some time, he decides to roll it over.

The child rotates the ring with a angular velocity of 'ω' and places it on the ground. The child is happy to notice that his logic worked as the ring translates forward for some distance and then topples over instead of toppling at the start. 

1. Find the distance 'D' through which the ring translates and the time 't' for which the ring translates before toppling over.

2. If instead the child had re-moulded the thin ring into a thin disc of the same thickness and used it for his above experiment on the same ground with the same value of angular velocity, determine the distance 'd' through which it translates before toppling over. Is 'd' less than, equal to or greater than 'D' ?

? · Accepted Answer

Here’s a solution, but can I point out that there are 3 problems with the question.

1.  It is not clear to what μ relates.  It could be:
a)	the frictional force which prevents slipping during the rolling motion (so if μ =0, the ring would slide, not roll);
b)	the rolling friction (rolling resistance), which determines how much energy is converted to heat as the ring rolls along.
These are 2 completely different things.  I will assume b) is applicable.

2 The problem could be interpreted in 2 ways:
a)	a complex problem involving the precession of a gyroscope (as the wheel has angular momentum);
b)	A simple one where the wheel rolls, loses its kinetic energy due to rolling friction (μ) and eventually stops and falls over.
If you are studying mechanics/physics/maths at university level and are currently dealing with gyroscopic motion, a) is relevant and I can’t help.  But if b) is applicable, my solution is applicable.  In real life, the ring would 'wobble and fall over' before befire it stopped.

3 Note that the cross-sectional shape must be assumed to be rectangular.
_________________________________________
Moment of inertia of ring = ½ M (r1² + r2²)
= ½ M[(3πR)² + (4πR)²]
= (25/2)π²MR²

Initial rotational kinetic energy = ½Iω²
= (25/4)π²ω²MR²

Initial linear velocity, v = ωr
Initial translational kinetic energy = ½mv²
= ½Mω²R²

Total kinetic energy = (25/4)π²ω²MR² + ½Mω²R²
= ((25/4)π² + ½)MR²

Rolling frictional force = μ x normal reaction = μMg

Work due by friction stopping ring = kinetic energy lost
μMg D =  ((25/4)π² + ½)MR²
D = ((25/4)π² + ½)R²/( μg)

When it stops after distance D, it will topple over.
__________________

Since acceleration is constant (as frictional force is constant) the time taken to stop is
t = distance/ (average linear speed)
= D/ [(0+ ωr)/2]
= 2D/(ωr)
where D is as determined above.
__________________________________________________
2) To do question 2, determine the volume of ring and calculate the radius of the new disc and its moment of inertia.  Then repeat the above calculation.

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Intriguing problem from Rotational Mechanics ?

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