- #1
rst
- 9
- 0
Homework Statement
I'm designing a device for changing a load position from vertical to horizontal. It has a wheeled frame (2) which allows an operator to transport the load after reorientation.
The combined centre of gravity of a rotating platform (1) and the load moves forward due to rotation, which may cause the frame (and thus the entire device) to move back, if a friction force between wheels and surface is insufficient. Brakes are engaged during the reorientation, therefore we consider sliding friction, not rolling.
I want to make sure that the frame stays in place. My calculations include determining a force which would push it back if there was no friction (conservation of momentum). If it outweighs the frictional force, then the device will slide.
Given data:
masses m1 (platform+load), m2 (frame)
angular velocty ω
CoG rotation radius R
initial CoG angle α0 (the angle varies from α0 to α0 +90 deg)
coefficient of sliding friction (wheels to surface) μ
Homework Equations
##
F_2 = \frac{dp_2}{dt} \\
p_1 + p_2 = 0\\
F_{fmax} = μ \cdot F_n = μ (m_1 +m_2)g\\
##
The Attempt at a Solution
1) Device standing on a frictionless surface
##
V = \omega \cdot R\\
V_1 = V \cdot cos(\alpha_0 + \omega t)\\
m_1 (V_1 - V_2) – m_2 V_2 = 0\\
V_2 = \frac{m_1 V_1}{m_1 + m_2}\\
p_2 = m_2 V_2 = \frac{m_1 m_2 V_1}{m_1 + m_2} = \frac{m_1 m_2 V cos(\alpha_0 + \omega t)}{m_1 + m_2}\\
F_2 = \frac{dp_2}{dt} = - \frac{m_1 m_2 V \omega}{m_1 + m_2}sin(\alpha_0 + \omega t)\\
|F_{2max}| = \frac{m_1 m_2 V \omega}{m_1 + m_2} = \frac{m_1 m_2 \omega^2 R}{m_1 + m_2} \\
##
2) The device will not move back if the following condition is satisfied:
## |F_{2max}| < μ (m_1 +m_2)g##
Question: is my reasoning valid?