- #1

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!

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!

Last edited by a moderator: