# Mechanics (general motion of a rigid body)

by Incontro
Tags: body, mechanics, motion, rigid
 P: 4 1. The problem statement, all variables and given/known data The bicyclist applies brakes as he descends a track making an angle β with a horizontal direction. The front wheel of the bike locks and starts to slip along the incline. Find the magnitude of the decleration a that would cause a dangerous condition of tipping about the front wheel. The combined center of mass of the rider and bicycle is at G and the total moment of inertia of the bicyclist and the bicycle about the axis perpendicular to the plane of the paper throufg G is Ig. The coefficient of frictions between the wheel and the track is μ. 2. The attempt at a solution There's no friction in the front wheel, right? And when there's a tipping of the front wheel the friction on the back wheel and the normal force on the back wheel is zero? But now I'm only left with the weight of the bicycle and the bicyclist and the normal force on the front wheel, and these forces can't make the deceleration....?
 P: 460 There is friction in the front wheel. Why not? It is slipping.
 P: 4 ahh, thanks! So how do I continue? I've got the weight, and the normal and the friction force at the front wheel. mag=F-mgsinβ Ma=mah=mghsinβ-mgbcosβ N=mgcosβ ??
P: 460
Mechanics (general motion of a rigid body)

 Quote by Incontro Ma=mah=mghsinβ-mgbcosβ
What is this? Are you equating mass with torque. about which point are you taking torque
 P: 4 Oh, my fault. I'm taking torque and around the point where the front wheel meet the ground. I'm then equating mah (the total force on the body times the length h between mass centre and A) with the torque on A.
 P: 460 What is your next attempt?
 P: 4 Mg=Igtheta(doubledot)=Fh-Nb=N((Mu)h-b) I want theta(doubledot) B is where the back wheel meet the ground. aB=aex-theta(doubledot)(b+c)ey ?

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