RBD BYU 7

We analyze the box in the rotating frame of the wheel.
Radial Distance: The rod is attached at the midpoint of the radius, so \( r = R/2 \).
Pseudo Force: Acts radially outward (Centrifugal). \[ F_p = m \omega^2 \left( \frac{R}{2} \right) \]

At an angle \( \theta \) with the horizontal:

• Vertical Component (Upwards): \( F_{py} = F_p \sin\theta = m \omega^2 \frac{R}{2} \sin\theta \)
  This component opposes gravity, reducing the normal force.
• Horizontal Component (Sideways): \( F_{px} = F_p \cos\theta = m \omega^2 \frac{R}{2} \cos\theta \)
  This component tries to slide the box along the rod. Friction must oppose this.

Vertical (Normal Force): \[ N + F_{py} = mg \implies N = mg – m \omega^2 \frac{R}{2} \sin\theta \]

Horizontal (Friction): \[ f = F_{px} = m \omega^2 \frac{R}{2} \cos\theta \]

Slip occurs when \( f > \mu N \). The limit is: \[ m \omega^2 \frac{R}{2} \cos\theta = \mu \left( mg – m \omega^2 \frac{R}{2} \sin\theta \right) \] \[ \omega^2 \frac{R}{2} (\cos\theta + \mu \sin\theta) = \mu g \] To find the minimum speed where slip implies finding the angle \( \theta \) that maximizes the “push” relative to the “grip”. Max value of \( (\cos\theta + \mu \sin\theta) = \sqrt{1+\mu^2} \).

$$ \omega^2 \frac{R}{2} \sqrt{1+\mu^2} = \mu g $$ $$ \omega = \sqrt{ \frac{2\mu g}{R\sqrt{1+\mu^2}} } $$

Linear velocity \( v = R\omega \):

$$ v = \sqrt{ \frac{2\mu g R}{\sqrt{1+\mu^2}} } $$