Register to reply

Mechanics (general motion of a rigid body)

by Incontro
Tags: body, mechanics, motion, rigid
Share this thread:
Incontro
#1
Mar12-11, 05:46 AM
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....?
Phys.Org News Partner Science news on Phys.org
Scientists develop 'electronic nose' for rapid detection of C. diff infection
Why plants in the office make us more productive
Tesla Motors dealing as states play factory poker
ashishsinghal
#2
Mar12-11, 06:37 AM
P: 460
There is friction in the front wheel. Why not? It is slipping.
Incontro
#3
Mar12-11, 11:48 AM
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 ??

ashishsinghal
#4
Mar12-11, 08:58 PM
P: 460
Mechanics (general motion of a rigid body)

Quote Quote by Incontro View Post

Ma=mah=mghsinβ-mgbcosβ
What is this? Are you equating mass with torque. about which point are you taking torque
Incontro
#5
Mar13-11, 03:56 AM
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.
ashishsinghal
#6
Mar13-11, 10:05 AM
P: 460
What is your next attempt?
Incontro
#7
Mar13-11, 11:48 AM
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 ?


Register to reply

Related Discussions
Rigid body motion Introductory Physics Homework 3
Classical mechanics: motion of a rigid body Classical Physics 0
Mechanics Book on 3-D Rigid Body Motion Science & Math Textbooks 2
Rigid body mechanics on the bus General Physics 2
What is rigid body motion? Classical Physics 1