CM M03 2

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The problem of a sphere rolling without slipping directly down an inclined plane is an old chestnut of freshman physics. In this problem, you are asked to examine more general motions of a sphere rolling without slipping on inclined and rotating planes.

The state of motion of a sphere of mass M is specified by giving its position d\vec{R}/dt with respect to a fixed point and its angular velocity \vec{\omega} with respect to its center of mass. The rolling without slipping constraint is usually expressed as d\vec{R}/dt=a\vec{\omega}\times\hat{n} where a is the sphere's radius and \hat{n} is the unit vector normal to the plane on which the sphere is rolling. There is a constraint force \vec{f} which acts in the plane to enforce the constraint and this constraint force has to be etaken into account in the equations for acceleration of the center of mass and for time rate of change of the angular momentum I\vec{\omega} about the center of mass.

a. Suppose that the plane is inclined to the horizontal at an angle θ in the earth's gravitational field. Show that you can eliminate the constraint force to find an equation for the center of mass alone and show that the center of mass experiences an acceleration of \frac{5}{7}g\sin\theta along the 'downhill' direction in the plane. Hence, the trajectries of this rolling sphere are parabolae.

b. Now consider that the plane is not inclined, but is rotating with angular velocity \vec{\Omega} about the vertical axis. The most important modification to the calculation you have just done is that the condition for rolling without slipping changes. Use the changed condition in your previous analysis to show that the free center of mass executes circular motion in the horizontal inertial frame with frequency \frac{2}{7}\Omega! In other words, the rolling sphere executes Lamor orbits, just like a charged particle in a uniform magnetic field!

For reference note that the moment of inertia of a homogenous sphere is \frac{2}{5}Ma^{2}.

Taking the time derivative of the rolling without slipping condition:

\frac{d^{2}\vec{R}}{dt^{2}}=a\frac{d\vec{\omega}}{dt}\times\hat{n}

The only torque on the ball is the force to enforce rolling without friction:

I\dot{\vec{\omega}}=a\hat{r}\times\vec{f}

Rearranging:

\frac{d\vec{\omega}}{dt}=\frac{a}{I}\hat{r}\times\vec{f}

Plugging this in:

\frac{d^{2}\vec{R}}{dt^{2}}=\frac{a^{2}}{I}\hat{r}\times\vec{f}\times\hat{n}

Since \hat{r}, so that \hat{r}\times\vec{f}\times\hat{n}=-\vec{f}:

\vec{f}=-\frac{I}{a^{2}}\frac{d^{2}\vec{R}}{dt^{2}}

Writing the force equation in the downhill direction:

m\frac{d^{2}R}{dt^{2}}=mg\sin\theta-\frac{I}{a^{2}}\frac{d^{2}R}{dt^{2}}

For a sphere I=\frac{2}{5}ma^{2}:

m\frac{d^{2}R}{dt^{2}}=mg\sin\theta-\frac{2}{5}m\frac{d^{2}R}{dt^{2}} \frac{7}{5}\frac{d^{2}R}{dt^{2}}=g\sin\theta

So we finally get:

\frac{d^{2}R}{dt^{2}}=\frac{5}{7}g\sin\theta

We change the condition for rolling without slipping to be:

\frac{d\vec{R}}{dt}+\vec{\Omega}\times\vec{R}=a\vec{\omega}\times\hat{n}

to take into account the motion of the ground. Taking the time derivative:

\frac{d^{2}\vec{R}}{dt^{2}}+\vec{\Omega}\times\frac{d\vec{R}}{dt}=a\frac{d\vec{\omega}}{dt}\times\hat{n}

\vec{f} will be defined in the same way as before:

\frac{d^{2}\vec{R}}{dt^{2}}+\left(\vec{\Omega}\times\frac{d\vec{R}}{dt}\right)=-\frac{a^{2}}{I}\vec{f}

So that in the force equation:

m\frac{d^{2}\vec{R}}{dt^{2}}=-\frac{2}{5}m\frac{d^{2}\vec{R}}{dt^{2}}+-\frac{2}{5}m\left(\vec{\Omega}\times\frac{d\vec{R}}{dt}\right)

or:

\frac{7}{5}\frac{d^{2}\vec{R}}{dt^{2}}=-\frac{2}{5}\left(\vec{\Omega}\times\frac{d\vec{R}}{dt}\right)

Since \vec{\Omega} is perpendicular to the plane:

\frac{d^{2}\vec{R}}{dt^{2}}=-\frac{2}{7}\Omega\left(\hat{z}\times\frac{d\vec{R}}{dt}\right)

This is the equation of motion for a circle in the plane, moving with frequency \frac{2}{7}\Omega!

This page was recovered in October 2009 from the Plasmagicians page on Prelim_M03_Mech2 dated 23:15, 1 January 2006.

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