It may seem counterintuitive, but the correct answer is that the frozen cylinder will roll slower, and so will take longer to roll down the inclined plane than the liquid-filled cylinder.

What's at play here?

The basic idea here is the balance of potential and kinetic energy. The cylinder starts out from rest, and any speed it picks up comes from Earth's gravitational field doing work on it. When a block slides down a plane, all the work goes into pulling that block down the plane.

When a cylinder rolls, the situation is different. Not only does energy go into the overall motion of the cylinder down the plane, but energy also goes into spinning the cylinder.

Think about it like this: we know that it takes work to pedal a bicycle up to speed. But even if we life the back tire off the ground, it still takes energy to spin up the back wheel, even if the bike doesn't move.

What does this mean for the cylinders?

In the question, an important assumption is given that when frozen, the ice does not move relative to the cylinder. This means we can treat the frozen cylinder as a rigid solid: in order for the cylinder to rotate, all of its mass must also be rotating at the same angular velocity. In other words, work is done not only to move the cylinder, but also to spin it.

If the ice melts, the cylinder is no longer a rigid solid, and the liquid inside can remain mostly still while the much lighter cylinder rotates around it. Since less gravitational potential energy is spent rotating the mass of the water, more will end up as translational kinetic energy.

A general lesson

For the rigid frozen cylinder, we say that it has a higher moment of inertia ; it takes more energy to get it moving, and keep it accelerating as it rolls. Consider the energy it takes to roll a hoola-hoop compared to a thick tire of the same radius -- a push that would send a hoola hoop rolling down the street might only budge a tire a few feet. The acceleration of gravity on the cylinder is the same between the two cases, so the greater moment of inertia of the frozen cylinder will result in a slower rotation.

A parting note

Several comments have focused on the movement of the water dragged into motion by the rotating cylinder and how it might dissipate energy due to friction. As the cylinder rotates faster, the liquid water inside will be driven into more turbulent (and dissipative) velocity profiles. Does this energy dissipation slow the roll more than the higher moment of inertia of the frozen cylinder? We decided to test this ourselves, so check out the video below featuring special guest @Josh Silverman !

Relevant wiki: Rotational Kinetic Energy - Translational Kinetic Energy

In the liquid state the potential energy of the water is basically transformed only to translational kinetic energy. In the solid state the potential energy is transformed into translational and rotational energy, since some energy goes into rotation, translational energy is reduced and the time to reach the bottom of the ramp is greater for the ice.

This is sort of an intuitive solution more than a scientific one: When water is in its liquid state, as the cylinder rolls down, water is going to be pushed to the back of the cylinder. If you've done this before, you might notice that when you have a cup of water and you move it suddenly in one direction, the water pushes in the opposite direction. The same will happen to the cylinder, and this opposing force will slow down the cylinder. In contrast, when it's frozen, it's not going to move around like this and push around on the cylinder.

Ice is less dense than water because as water cools and becomes a solid (freezes), hydrogen bonds form between the water molecules. In the liquid phase of water, the molecules "snuggle up" to each other in the fluid. But as water goes solid, the hydrogen bonds dictate that the molecules will have to "stop snuggling" and move apart a bit as those hydrogen bonds set up spacing in the now-solid molecules. Ice has become less dense than the water that it formed from because the hydrogen bonds, which begin forming at just above 0 °C, force the molecules apart a bit to form the solid (ice) matrix.

secondly,, in the solid stage-the whole energy will be translated into-motion energy and turning energy

while in the water stage the whole energy will be translated into- motional kinetic energy.

thus,frozen water will be faster.

I had a different reasoning.

Ice has lower density (0.9167 gm/cc) than water (0.9998 gm/cc). Given that both the cylinders are of same volume, it can be inferred that the cylinder with water would be heavier and hence will be faster.

Liquid water can roll around inside the cylinder, so the moment of inertia resembles that of a hollow cylinder whereas with the ice the moment of inertia is that of a solid cylinder, which is smaller, so the cylinder rolls down faster with ice.

This problem has many aspects.

Simplest model : The ice is frozen stuck to the inside of the cylinder. Therefore, when the cylinder rolls down the ramp, the cylindrical chunk of ice must also spin at the same rate. Strong forces between the cylinder and the ice ensure this. These forces will slow down the cylinder and speed up the ice. In contrast, the melted water can flow independent of the cylinder. There are no strong forces between the cylinder and the water; therefore the cylinder is not slowed down as much.

Or: at any given height $h$ , the kinetic energy of the cylinder and its content is $K = E - U = mg(h_0 - h)$ . In the first case, a good part of this energy is in the rotation of the chunk of ice. In the second case, little energy is needed for the relative motion of the water.

Conclusion: $\boxed{\text{the cylinder with frozen water takes longer}}$ .

Problems with the frozen water : We assumed that the ice is frozen to the cylinder, so that $\omega_{\text{cyl}} = \omega_{\text{ice}}$ . If the ice is not stuck to the inside of the cylinder, the entire process can happen without the ice rotating. The cylinder with the frozen water will then go almost as fast, or perhaps even slightly faster, than the one with melted water.

Problems with fluid friction : Suppose the second cylinder is entirely filled with liquid water. As the cylinder starts spinning, water near its inner surface will experience drag force (due to viscosity). Thus the water will start spinning, just as in the case of the ice, taking away from the motion of the cylinder. Friction between fluid layers will eventually cause all the water to spin, much as in the case of the ice. It will take longer than in the case of ice, but the result is similar. This is especially true if the angle of the incline is small.

One might argue that the liquid water near the center will always have a smaller rotational speed than the cylinder itself, and therefore overall less rotational kinetic energy than the spinning chunk of ice. This is true. However, an additional effect slows the cylinder with the liquid even further: since this friction is not static friction but kinetic friction, it will dissipate kinetic energy. While the spinning water may end up with less kinetic energy than the spinning ice, it may have generated fair amounts of thermal energy, so that the cylinder with liquid may be slower after all.

Problems with the melting : As many have indicated, the melting will reduce the volume of the water by about 8% if the pressure remains constant. This will create a vacuum in the cylinder, which may retain its shape or shrink. If the cylinder shrinks without changing its overall shape, this does not really affect the time its takes to roll down the ramp; this time depends only on the mass distribution (or on $\kappa/r$ , where $\kappa$ is the radius of gyration and $r$ the radius of the rolling motion).

Another problem with the melting is that water will slosh around. There may be waves at the surface and collisions between water drops and the inside of the cylinder, all of which lead to quick dissipation of mechanical energy. The overall effect is further slowing of the cylinder with liquid.

Conclusion: Especially on a small-angle incline, issues of friction and melting may result in $\boxed{\text{the cylinder with liquid water taking longer}}$ .

Let

• $M$ be the mass of the cylinder and water
• $I$ be the the moment of inertia
• $ω$ be the angular velocity
• $R$ be the radius
• $h$ be the height of the center of the cylinder off the ground
• $θ$ be the angle of the ramp
• $v$ be the velocity vector of the center of the cylinder
• $T$ kinetic energy
• $V$ potential energy
• $L$ Lagrangian, $T - V$

Assuming the cylinder isn't slipping, we have $|v| = R ω$

Thus $\dot{h} = -|v| sin θ = -R ω \sin θ$

$T = \frac{1}{2} M |v|^2 + \frac{1}{2}Iω^2 = \frac{1}{2}(MR^2 + I)ω^2 = \frac{1}{2}(MR^2 + I) (\frac{-1}{R \sin θ} \dot{h} )^2$

$V = Mgh$

Now we'll use the Euler-Lagrange equation $\frac{\partial L}{\partial h} = \frac{d}{dt}\frac{\partial L}{\partial \dot{h}}$ to get the equation of motion:

$\frac{\partial L}{\partial h} = -Mg$

$\frac{d}{dt}\frac{\partial L}{\partial \dot{h}} =\frac{ (MR^2 + I)}{R^2 \sin^2 θ} \ddot{h}$

Thus

$\ddot{h} = - \frac{MR^2 }{MR^2 + I} g \sin ^2 θ$

Now notice what happens to this equation as we change $I$ . If $I = 0$ it reduces to $\ddot{h} = -g \sin ^2 θ$ . As $I$ gets larger, $\ddot{h}$ gets smaller. If the water is melted, at least in the ideal case, we can assume the cylinder just slides right past the water without friction, so the water doesn't rotate or contribute to $I$ . In the frozen case, the water does rotate. So in the frozen case $I$ is larger, so according to the equation above, the frozen case accelerates slower.

The ice would expand and break the container in the translation from water to ice

When me mix a slurry of fruit pulp in a mixy or juicer you might notice after running it and opening the lid most of the pulp sticks to the sides with the center being empty. This is due to centrifugal force.

If we compare the moment of inertia of a hollow cylinder to a solid cylinder, the hollow cylinder has a lower moment which causes it to roll faster down an incline.

how about using energy level to describe kinetic energy?

• frozen water has less initial energy level (Potential energy + Small Internal Energy)
• Melt water has more initial energy level (Potential energy + large internal Energy(absorb from the air) )

Read carefully: the question was is it faster when frozen or AFTER it melts. So the frozen one starts rolling instantly but the other one has to wait till it melts and starts then.

Since volume of ice is greater than water for a given mass in addition to the weight of the barrel which will act perpendicular to the ground, there will be a small force acting perpendicular to the plank on which the barrel rolls.. by resolving this I think we see that the force from the weight of the barrel increases and additionally a force opposite to the downward motion is also created... this is how I visualized it... I may be wrong... need your comments... Thanks

My thought was that since water is more dense than ice, there's more mass in the liquid filled cylinder as opposed to the ice one. That means that the liquid cylinder would accelerate slower due to inertia.

Since the melted state is a cylinder “completely filled” with water, the cylinder would have to expand when frozen. So, in addition to the other explanations, we have to consider that the frozen state would introduce some wobble, further slowing it down.

In the liquid state of the water it would have less gravitational potential energy than the frozen water,because when water is frozen it is much denser causing it to roll faster

I am assuming that the volume of water is the same in both cylinders (and hence the weight) because the body is strong enough to contain the expanded frozen liquid.

In the frozen state the cylinder acts as a solid body. In the melted state energy is transferred to rotational fluid movement as well as kinetic (down the slope) energy. With more energy transferred as a solid body the frozen body will roll down more efficiently.

Are there any formulas for the calculation? What about other shapes? I found these links: http://farside.ph.utexas.edu/teaching/301/lectures/node108.html https://m.youtube.com/watch?v=cPMSMDSY-rc https://isaacphysics.org/questions/rolling_objects

Solution

So, the basic idea is potential energy (mass m ) gets converted to kinetic (translational and rotational) energy