# Rotation of fluid filled cylinders [closed]

Given two cylinders of same mass , one completely made out of solid metal and the other a hollow one which has been filled with a viscous fluid say oil.

Both have the same mass.Now if they are placed on an inclined plane just so that a force acts on it, will their parameters of motion such as acceleration be different? Assume that both of them undergo pure rolling.

So the Question is will they have the same accelerations?

If no what is the cause for such an observation?

• @Gert , have you actually witnessed the situation that you describe below? I wouldn't assume that "I" for the fluid filled cylinder starts at zero. There is a lot of drag between the wall of the cylinder and the fluid next to it. This drag will strongly resist rotation. I don't know if the situations are even remotely similar, but try spinning a raw egg and a hard-boiled egg. You will notice that the raw egg does not spin for very long due to the huge drag effects from the egg white inside the shell. – David White Dec 6 '15 at 18:24
• @DavidWhite: zero $I$ was clearly an illustrative simplification and only valid for low viscosity fluids where shear forces between wall and outer fluid film are very small. But even for higher viscosity fluids there will always be lag between the $\omega$ of the shell and of the fluid itself (and between the outer layers of fluid and inner ones). The egg is a case of high viscosity and one where you stop applying external torque. – Gert Dec 6 '15 at 18:32
• I'll have to work up a demo for this case so I can see it for myself. – David White Dec 6 '15 at 18:34

So the Question is will they have the same accelerations?

If no what is the cause for such an observation?

No, assuming all other things equal (including radius $R$ of both cylinders), the fluid filled one will accelerate faster.

Consider the following diagram of a cylinder on an inclined plane, viewed sideways: On such a cylinder acts an accelerating force of (with Newton):

$$mg\sin\theta=ma,$$

which causes acceleration $a$ parallel to the plane.

But the cylinder also undergoes rolling without slipping, which is caused by the friction force between the cylinder and the plane. This friction force exerts a torque $\tau$ that causes an angular acceleration $\alpha$ acc.:

$$\tau=I\alpha,$$

where $I$ is the inertial moment of the cylinder.

We need not consider ourselves with the actual magnitude of $\tau$ because of the 'no-slippage' condition, which in mathematical terms means, at all times:

$$v=\omega R,$$

with $v$ the translational speed, $\omega$ the angular speed and $R$ the radius of the cylinder.

Now let's look at the energy conservation equation; assuming no losses then potential energy $U$ is constantly converted to kinetic energy $K$ with the sum $T$ always remaining constant:

$$T=K+U$$

Potential energy here equals:

$$U=mgy,$$

where $y$ is the height of the object above the horizontal plane.

$K$ has two components, one due to translation and one due to rotation:

$$K=\frac{mv^2}{2}+\frac{I\omega^2}{2}=\frac{mv^2}{2}+\frac{Iv^2}{2R^2}$$

Or:

$$K=\bigg(\frac{m}{2}+\frac{I}{2R^2}\bigg)v^2$$

For ease of notation I'll set: $\beta=\bigg(\frac{m}{2}+\frac{I}{2R^2}\bigg)$.

We can now write:

$$T=mgy+\beta v^2$$

If we define an $x$-axis parallel to the inclined plane, then $y=x\sin\theta$ and we get:

$$T=mgx\sin\theta+\beta v^2$$

Deriving both sides to time ($\frac{d}{dt}$) and knowing that $T$ is a constant we get:

$$mg\sin\theta v+2\beta v\frac{dv}{dt}=0$$

$$mg\sin\theta+2\beta a=0$$

So:

$$\large{a=-\frac{mg\sin\theta}{2\beta}}$$

Note that the negative sign is due to the choice of $x$-axis: acceleration is in the opposite direction of the $+x$-direction.

And now we can see why the fluid filled cylinder will accelerate faster than an equivalent solid cylinder.

For a solid cylinder the inertial moment is given by:

$$I=\frac{m}{12}(3R^2+H^2),$$

where $H$ is the height of the cylinder. So we need to insert that value into $\beta$ to get $a$.

Now take the case of a cylinder filled with a lowly viscous fluid. As there is no torque to make the 'fluid cylinder' rotate, it simply doesn't (or only very slightly). So while the cylinder rotates, the fluid itself doesn't. As a crude approximation we can say that $I \approx 0$ and then:

$$a=-g\sin\theta,$$

Which is much larger than the case of the solid cylinder.

For more viscous fluids the 'fluid cylinder' will start rotating quicker and $I > 0$ so $a$ will be reduced then.

In all cases, given enough time, the 'fluid cylinder' will eventually rotate at the same speed as its hollow shell, in which case the situation is reduced to that of the equivalent solid cylinder. We can even conclude that the value of $a$ will gradually decrease as the fluid filled cylinder continues to roll and its fluid core comes up to the shell's angular speed $\omega$.

• Ok got that. But lets just say that there exist friction between the fluid and inner wall of cylinder . Will that affect the cylinder? – hybrid1999 Dec 7 '15 at 8:58
• @HrushikeshBodas: Hi. YES. Friction between the wall (so-called viscous shear stress) and the outer layer of liquid, as well as shear stress between the liquid layers is the torque that does make the fluid rotate inside the shell. The simple corollary is that two identical cylinders, one with a low viscosity, one with a high viscosity fluid, the former will accelerate faster than the second one. And fill the cylinder with a very high viscosity fluid and it will behave almost like the solid cylinder. – Gert Dec 7 '15 at 16:05
• Incidentally, you can check all this quite simply as follows. Take two identical bottles filled 90 % with water and freeze one of them. Then let them roll off a slightly inclined table, in identical conditions. – Gert Dec 7 '15 at 16:08
• Yes got that thanx – hybrid1999 Dec 9 '15 at 15:30