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vmc7628 · Apr 3, 2013, 12:07 PM

A solid marble of radius a rolls without slipping back and forth in the bottom of a
hemispherical bowl of radius R. Show that, for small oscillations, the angular frequency of
oscillations is
omega knot = the square root of 5 g / 7R -a

ebaines · Apr 3, 2013, 01:08 PM

You will need to set up a differential equation that relates the linear + rotational acceleration of the marble to the gravitational force and resulting torque being applied to the marble as it rolls. To develop a single DE that combines both rotational and linear displacements you will have to relate the angular rotation of the marble to its angular displacement up the side of the bowl. Please show us your attempt and we'll help if you get stuck. Couple of tricks to remember: (1) the distance from the center of the hemispherical bowl to the center of the marble is (R-a) units, and (2 ) when considering the angular displacement $\theta$ of the marble from the bottom of the bowl for small values of displacement $\sin(\theta) = \theta$ .

vmc7628 · Apr 6, 2013, 09:59 AM

Originally Posted by ebaines

You will need to set up a differential equation that relates the linear + rotational acceleration of the marble to the gravitational force and resulting torque being applied to the marble as it rolls. To develop a single DE that combines both rotational and linear displacements you will have to relate the angular rotation of the marble to its angular displacement up the side of the bowl. Please show us your attempt and we'll help if you get stuck. Couple of tricks to remember: (1) the distance from the center of the hemispherical bowl to the center of the marble is (R-a) units, and (2 ) when considering the angular displacement $\theta$ of the marble from the bottom of the bowl for small values of displacement $\sin(\theta) = \theta$ .

Could you be a little more specific with your assistants please. I do not want you to give me too much but as a struggling physics student I could use a little more if possible. I have been reading your reply for the past 2 days and attempting to construct an equation to solve for omega knot. I have not been able to produce a equation such as you described I have only small pieces from notes that I think will help such as inertia of a solid ball 2m^2/5. Check out the only paper of the 20 or 30 that I believe has any useful information on it.

ebaines · Apr 6, 2013, 07:51 PM

Think about the marble with displacement angle $\theta$ from the midpoint of the hemispherical bowl - it has a force of gravity acting in the direction parallel to the surface of the bowl of $F= mg \sin \theta$ , which is equal to mass times acceleration. That acceleration is equal to $r\ddot \theta$ , where r is the radius of the circular motion of the marble about the center of the hemisphere, which is the distance R-a. The direction of that force of gravity is opposite to the positive direction $\theta$ , so:

$m(R-a) \ddot \theta = -mg \sin \theta$

Note: the notation $\ddot \theta$ means $\frac {d^2 \theta}{dt^2}$

This accounts for the linear motion of the marble (as if it was sliding - BTW this ought to look familiar to you as it's also the equation of motion for simple pendulum), but you must also consider the fact that the marble rolls. The torque acting on the marble is $T = -amg \sin \theta$ , and that torque equals $I \ddot \alpha$ where $\alpha$ is the rotational angle of the marble as it rolls along the inside surface of the hemisphere. So: $I \ddot \alpha = - amg \sin \theta$

Putting these together, including multiplying through the first equatuion by 'a' so that it's right jhand side is in agreement with the second equation's right hand side yields:

$I \ddot \alpha + m a(R-a) \ddot \theta = -amg \sin \theta$

or, if you take into account that $I = \frac 2 5 ma^2$ and for small angles $\sin \theta = \theta$ :

$\frac 2 5 ma^2 \ddot \alpha + ma(R-a) \ddot \theta + amg \sin \theta = 0$

Almost there. Now you need to come up with a way to express $\alpha$ as a function of $\theta$ , so that the above DE is in terms of $\theta$ only. I'll leave that to you. Once you get that resolved you have a simple DE that will have a solution of the form $\theta = A \sin (\omega t + \phi)$ , where 'A' is the amplitude of motion and $\phi$ is the phase angle. Values for 'A' and $\phi$ can be found from initial conditions if they were given to you, but for the purposes of this problem their values are immaterial since you are only trying to solve for $\omega$ .

Post back with your attempt at getting to the final answer for $\omega$ .