Do cylinders with different radii roll at different rates?

Question

If you have a cylinder with say r = 2, and its mass is uniformly distributed, would it get to the bottom of an inclined plane at a different time then a cylinder with r = 4? I think so, but I'm not totally sure.

Answer 1

No, homogeneous cylinders with different radii accelerate and roll down parallel to the inclined plane at the same rate. It's even independent of their masses as well, provided that both masses are distributed in the same relative way, that is, uniformly in this case. That's just the way that gravity works.

(Note that the use of the word "roll" can be ambiguous. I'm using "roll down" in the sense that it's certainly used colloquially, as in "the car rolled down the hill." Now literally, of course, the car itself didn't roll! Only its wheels were really "rolling." If "rolling" is instead meant as a synonym for "angularly rotating," then of course a larger cylinder will "roll" at a slower rate. The rate of such angular "rolling" for differently sized cylinders is simply inversely proportional to their radii. The angular "rolling rates" are connected through the relationship(s) w_1 r_1 = v = w_2 r_2, where v is the (same) speed with which they are moving down the plane at any given time or place --- see my appended dynamical treatment, below. *** )

I'm assuming you're saying that they both have a sufficient coefficient of friction so that they roll without slipping. In that case, as the forces involved are proportional to their masses, and their Moments of Inertia are proportional to their masses and the squares of their radii, you'll find that the same fractions of gravitational energy tapped by rolling down the slope will go into their separable kinetic energies of linear motion and rotation. That's ultimately how the times to roll down the same distance are the same.

You will get different times if you roll two cylinders down with different mass distributions, for example a solid, homogeneous cylinder and a hollow cylinder. That's ultimately because, in that case, different fractions of the tapped gravitational energy go into those two separate components of the total kinetic energy.

Live long and prosper.

*** P.S. O.K., enough of all these other random guesses that are being made without rigorous dynamical support. I decided I had to do the problem, and it's exactly as I first suggested. Consider an inclined plane at angle 'theta' to the horizontal. If the mass of a circular rolling body (cylinder or sphere) is M, its radius R, its Moment of Inertia about its centre ' k MR^2, ' and it rolls down the inclined plane from rest without slipping,with linear speed v and angular speed w when at a distance rolled of d from rest, then the following conditions hold:

1. v = wR (no-slip condition),

2. Linear K.E. = (1/2) M v^2,

3. Rotational K.E. = (1/2) k M R^2 w^2 = (1/2) k M v^2, so that:

4. Total K.E. = (1/2) (1 + k) M v^2, and:

5. Gravitational Potential energy released = M g d sin (theta)

By conservation of energy, the expressions in eqns. 4. and 5. are equal, therefore:

6. (1/2) (1 + k) M v^2 = M g d sin (theta).

Notice that not only has any reference to R, per se, disappeared, but M also divides out. (As I said above, it ALWAYS does in simple problems under gravity like this.) This leaves us with:

7. v^2 = 2 [1/(1 + k)] g d sin (theta).

Comparison with that summary expression known if not loved by generations of physics students:

8. " v^2 = 2 a d ",

shows us that ' a ' in this case is given by:

9. a = g [sin (theta) / (1 + k)].

Note that if k = 0 (NO rotational energy!) we recover the usual acceleration down the slope of g sin (theta).

Provided the relative mass distributions are equal (==> same k), the acceleration is independent of mass and radius. However, bodies with different k's will accelerate at different rates.

Recall that for a hollow cylinder, k = 1, so the acceleration is halved in comparison to the standard value for a freely sliding object. For a homogeneous cylindrical distribution, k = 1/2, so the acceleration is 2/3 of that standard. A uniform sphere has k = 2/5, if I recall correctly, so the acceleration is 5/7 of the standard.

Thus the answer to the original question is that two homogeneous cylinders of different radii will in fact roll down to the bottom of the inclined plane in exactly the same time.

Answer 2

I can go into a lot more detail if you want, but the short answer is no. Both solid uniform cylinders would reach the bottom of the inclined plane at the same time, assuming they roll without slipping.

Answer 3

It's a trick question. Both cylinders will travel the same speeds. Yes, they will reach the bottom at the same time. However, the larger cylinder will take less revolutions since it has a greater surface. Why? Because in one turn it travels a farther distance. So they will roll at a different rate, but arrive at the same time.

Answer 4

Charges start @ $500 to do other's schoolwork.

#1 A clue can be found in the circumference of each cylinder.

#2 Assumes that the cylinder's surfaces are identical, as are all other factors with exception of thier differing radii.

# 3 Remittance address to follow.

Answer 5

If there is no friction, the weight has no effect. In the presence of friction, it takes more energy to stop a heavier weight so it will travel further. However, if friction is increased by weight, that will tend to shorten the rolling distance. On a hard surface, rolling friction is small, but on a rubber-like surface, rolling friction will increase with weight. Then you would have to know the values of friction vs weight to get an answer for a particular situation.

Answer 6

the small one goes faster becauase of th low inetia.inertia is the resistence for movement.I=mr^2.since these are uniformly distributed cylinder with large radius will have a largr inertia.so it will roll slowly

Answer 7

yes a cylinder with a small ratio would roll faster that one with a larger ratio

Answer 8

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Answer 9

no