Ball rolling in a magnetic field

[etotheipi]

Homework Statement

A ball of radius R, mass M and uniform charge Q is set rolling with its centre of mass initially moving at a speed $V$. The magnetic field is perpendicular to the surface, which is flat and has a large enough coefficient of static friction to prevent slippage. Determine the trajectory of the ball.

Relevant Equations

N/A

I first found the Lorentz force on the ball as a whole$$\vec{F}_m = \iiint_V \rho(\omega \times \vec{r} + \vec{V})\times \vec{B} dV = \rho \vec{\omega} \times \left( \iint_V \vec{r} dV \right) \times \vec{B} + \rho \iiint_V \vec{V} \times \vec{B} dV = Q\vec{V} \times \vec{B}$$due to the spherical symmetry of the ball, which demonstrates that the magnetic force on the ball is equivalent to that on a point charge at the centre of mass. For the rest I will take the $\hat{z}$ direction as upwards.

Euler's law of motion applied to the ball, with the force of static friction $\vec{F}_f$ and the magnetic force $\vec{F}_m$, yields$$Q\vec{V} \times \vec{B} + \vec{F}_f = m\vec{a}_{cm}$$Then I also took torques about the centre of mass, noting that the magnetic force causes a torque (due to the presence of a magnetic dipole caused by induced currents in the ball) which is $\vec{\tau}_m = \frac{Q}{2m} \vec{L}_{cm} \times \vec{B} = \frac{Q}{5}R^2 \vec{\omega} \times \vec{B}$ where $\vec{L}_{cm} = \frac{2}{5}mR^2 \vec{\omega}$ is the angular momentum about the centre of mass:$$\sum \vec{\tau}_{cm} = (-R\hat{z})\times \vec{F}_f + \frac{Q}{5}R^2 \vec{\omega} \times \vec{B} = \frac{2}{5}mR^2 \vec{\alpha}$$Now, I define the unit vector $\hat{n}$ in the direction of the frictional force such that $\vec{F}_f = \mu m g \hat{n}$ and write $\vec{\alpha}$ as $$\vec{\alpha} = -\frac{5}{2mR} \hat{z} \times (\mu mg) \hat{n} + \frac{Q}{2m}\vec{\omega} \times \vec{B}$$I can now use the rolling condition, $\vec{a}_{cm} = R\vec{\alpha} \times \hat{z}$, to write down$$\vec{a}_{cm} = -\frac{5\mu g}{2}\hat{n} + \frac{QR}{2m} \omega B \hat{n}$$This demonstrates that the friction force is oppositely directed to the magnetic force. Finally I can insert this into the Euler's law equation to obtain$$-QVB\hat{n} + \mu m g \hat{n} = -\frac{5\mu mg}{2} \hat{n} + \frac{QVB}{2}\hat{n}$$This means that we can rewrite $\vec{a}_{cm} = -\frac{4}{7} \frac{QVB}{m} \hat{n} = -r \Omega^2 \hat{n}$. However, the solution manual notes that $\Omega = \frac{6QB}{7m}$ and $r = \frac{7mV}{6QB}$ for the resulting circular motion, which isn't what I get. I'm out on $\Omega$ by a factor of $\frac{2}{3}$. So I wondered whether anyone could see where I have gone wrong? Thanks!

 

[TSny]

I think your work is basically good. There is no need to write the friction force in terms of the coefficient of friction. In fact, the friction is static friction. So, it is not necessarily true that $f = \mu mg$. In general, $f \leq \mu mg$.

Now, I define the unit vector $\hat{n}$ in the direction of the frictional force such that $\vec{F}_f = \mu m g \hat{n}$ and write $\vec{\alpha}$ as $$\vec{\alpha} = -\frac{5}{2mR} \hat{z} \times (\mu mg) \hat{n} + \frac{Q}{2m}\vec{\omega} \times \vec{B}$$I can now use the rolling condition, $\vec{a}_{cm} = R\vec{\alpha} \times \hat{z}$, to write down$$\vec{a}_{cm} = -\frac{5\mu g}{2}\hat{n} + \frac{QR}{2m} \omega B \hat{n}$$ This demonstrates that the friction force is oppositely directed to the magnetic force.

I believe there is an inconsistency in your signs here. Note that $(\vec \omega \times \vec B) \times \hat z$ points in the direction of the magnetic force. But you took $(\vec \omega \times \vec B) \times \hat z$ to be in the direction of $\hat n$. You defined $\hat n$ as the direction of the friction force and concluded that the direction of the friction force is opposite to the direction of the magnetic force. So, something's wrong.

 

[etotheipi]

I think your work is basically good. There is no need to write the friction force in terms of the coefficient of friction. In fact, the friction is static friction. So, it is not necessarily true that $f = \mu mg$. In general, $f \leq \mu mg$.

I believe there is an inconsistency in your signs here. Note that $(\vec \omega \times \vec B) \times \hat z$ points in the direction of the magnetic force. But you took $(\vec \omega \times \vec B) \times \hat z$ to be in the direction of $\hat n$. You defined $\hat n$ as the direction of the friction force and concluded that the direction of the friction force is opposite to the direction of the magnetic force. So, something's wrong.

You're a lifesaver, I've been stressed out about this for the whole evening .

You're completely right, $\frac{Q}{2m}(\vec{\omega} \times \vec{B}) \times \hat{z} = -\frac{QVB}{2m} \hat{n}$, and if I substitute that through I get that, along with $V = r\Omega$,$$-\frac{6}{7}\frac{QVB}{m} = -r\Omega^2$$ $$\Omega = \frac{6}{7}\frac{QB}{m}$$which is what is given in the solutions.

Thank you ☺