A Sphere Oscillating in the Bottom of a Cylinder

In summary, the question asks for the frequency of small oscillations of a small ball with radius r and uniform density that rolls without slipping near the bottom of a fixed cylinder of radius R. The solution involves calculating the forces on the center of mass of the ball, which include friction and gravity. The differential equation \ddot{\theta}+\frac{5g}{7R}\theta =0 is derived and solved to find the frequency \omega = \sqrt{\frac{5g}{7R}}. There may be confusion regarding the work done by friction and the use of a non-inertial frame, but these can be explained by considering the definition of work and the relative nature of frames of reference.
  • #1
hbweb500
41
1

Homework Statement



"A small ball with radius r and uniform density rolls without slipping near the bottom of a fixed cylinder of radius R. What is the frequency of small oscillations, assuming r<<R?"

http://img89.imageshack.us/img89/8614/helpwu8.png

Homework Equations



[tex]
F=ma
[/tex]

[tex]
\tau=I\alpha \\
[/tex]

[tex]
I=\frac{2}{5}mr^2 \\
[/tex]

[tex]
a = \ddot{\theta} R= \alpha r
[/tex]

The Attempt at a Solution



The solution I can get, I just have a few questions about the details.

First, writing out the forces on the CM of the ball: the force of friction points in the same direction as the velocity vector, due to the non-slipping constraint. The magnitude of the ball's velocity is decreasing the greater the angle between it and the bottom of the cylinder, and thus the angular frequency of its rotation must also decrease. Thus the friction force is opposite the rotation.

The other relevant force is obviously gravity, scaled by a sine of theta:

[tex]
F_{\hat{\theta}}= F_f - mg sin\theta = ma
[/tex]

And there is a also a torque equation. Relative to the CM of the ball, we have a torque exerted by friction. Thus:

[tex]
-r F_f = \frac{2}{5}mr^2 \alpha
[/tex]

Combining these two equations, and assuming that, for small theta, [itex]\theta\approx sin\theta[/itex], we get the differential equation:

[tex]
\ddot{\theta}+\frac{5g}{7R}\theta =0
[/tex]

[tex]
\omega = \sqrt{\frac{5g}{7R}}
[/tex]

What I can't understand...

1. The Work Done By Friction
The bottom of the ball is instantaneously at rest with respect to the bottom of the cylinder. The force of static friction, then, exerts a force over zero distance, and thus does no work. However, it also is responsible for the change in angular frequency. Without the friction, the ball would not rotate.

The rotational kinetic energy of the ball is changing due to the friction force, but the friction force does no work. I don't get this.

2. Non-inertial Frame?
The CM of the ball is accelerating, yet the torque is still calculated using it as an origin. How is this legal?

If it matters, this is problem 8.13 in Morin's Introduction to Classical Mechanics.

Thanks for your help!
 
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  • #2
If you remember back to centripetal motion, recall that the centripetal force does no work as well. Since the force did not move the sphere, it just merely changes the direction of the sphere. Therefore it didn't do any "real" work.

I guess it really goes back to how work is define. f (dot) x.

I have a feeling that I am not making a lot of sense


Edit:
oops didn't see part 2.

Well, frame of reference are relative. You can take any frame of reference to calculate physics (if you can, some frame of reference are a pain in the latus rectum). The Earth is in motion (with respect to the sun) everyday, but we calculate most of physics without considering that motion.
 
  • #3
Yes, but a centripetal force does not change the magnitude of the velocity, thus there is no change is the total kinetic energy of the object it pulls on.

The friction is transferring some of the translational kinetic energy of the ball to rotational. I have a hunch that the total kinetic energy of the ball does not change, though, so the friction does no work.

But then again, in the angular direction, the friction force is in the f=ma equation.

And I see what you mean about the non-inertial frame, but it just seems fishy to me. The size of the system of something on Earth is negligible compared to the Earth-Sun system, so that can be ignored. But taking the ball in the cylinder as a system, it seems that the effect of the non-inertial frame might be more noticeable.
 

1. What is the phenomenon of a sphere oscillating in the bottom of a cylinder?

The phenomenon of a sphere oscillating in the bottom of a cylinder refers to the motion of a sphere that is placed inside a cylindrical container and allowed to move freely. The sphere will move back and forth, or oscillate, due to the forces of gravity and buoyancy acting on it.

2. What factors affect the oscillation of the sphere in the cylinder?

The oscillation of the sphere in the cylinder can be affected by various factors, including the size and weight of the sphere, the depth and diameter of the cylinder, and the viscosity of the liquid inside the cylinder. Other factors such as external vibrations or disturbances can also affect the oscillation.

3. How is the period of oscillation determined for a sphere in a cylinder?

The period of oscillation for a sphere in a cylinder can be determined by the size and weight of the sphere, as well as the properties of the liquid inside the cylinder. The period is also affected by the depth and diameter of the cylinder, as well as any external factors that may influence the oscillation.

4. What is the relationship between the sphere's oscillation and the liquid's density in the cylinder?

The density of the liquid inside the cylinder can affect the oscillation of the sphere by changing the buoyant force acting on it. If the density of the liquid is increased, the buoyant force will also increase, causing the sphere to oscillate at a slower rate. Conversely, a decrease in liquid density will result in a faster oscillation.

5. How is the energy of the system conserved during the oscillation of the sphere in the cylinder?

The energy of the system is conserved during the oscillation of the sphere in the cylinder due to the principle of conservation of energy. As the sphere moves back and forth, it will experience changes in potential and kinetic energy, but the total energy of the system will remain constant. This is because the forces acting on the sphere, such as gravity and buoyancy, are conservative forces that do not dissipate energy.

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