MEC BYU 27

Minimum Current to Flip:

For the ring to flip, the magnetic torque must exceed the gravitational torque about the tangent pivot point.

$$ \tau_m = \tau_g $$ $$ (I_{min} \pi r^2) B = mg r $$ $$ I_{min} = \frac{mg}{\pi r B} $$

The applied current is $I = 2 I_{min} = \frac{2mg}{\pi r B}$. This implies $mg = \frac{1}{2} I \pi r B$.

Normal Reaction Calculation:

Consider the rotation about the pivot (edge of the ring). The net torque is:

$$ \tau_{net} = \tau_m – \tau_g = (I \pi r^2 B) – mg r = (2 mg r) – mg r = mg r $$

Angular acceleration $\alpha = \tau_{net} / I_{pivot}$. Moment of inertia about tangent is $\frac{3}{2}mr^2$.

$$ \alpha = \frac{mg r}{\frac{3}{2}mr^2} = \frac{2g}{3r} $$

The center of mass accelerates upwards tangentially: $a_{cm} = r \alpha = \frac{2g}{3}$.

Using Newton’s second law in vertical direction: $N – mg = m a_{cm}$.

$$ N = mg + m(\frac{2g}{3}) = \frac{5}{3} mg $$

Substitute $mg = \frac{1}{2} I \pi r B$: