Determine the coefficient of static friction between car and track

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Homework Help Overview

The original poster presents two distinct physics problems: one involving a car on a circular track experiencing static friction and another concerning the moment of inertia and rotational energy of a system of particles. The focus is on determining the coefficient of static friction in the first problem.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Some participants suggest equating forces to extract the coefficient of friction, while others inquire about the original poster's ideas and understanding of the problem. There is also a mention of weights for the second problem, indicating a need for clarification on the parameters involved.

Discussion Status

The discussion is ongoing, with participants providing hints and guidance on the first problem. There is a mix of responses, with some participants expressing confidence in the simplicity of the second problem, while the first problem remains less resolved.

Contextual Notes

Participants note the importance of considering the weights of the particles in the second problem, which may affect the calculations for moment of inertia and energy. The original poster seeks direction without explicit solutions, adhering to homework guidelines.

HurricaneH
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Any help will be aprreciated...

1. A car traveling on a flat circular track accelerates uniformly from rest with a tangential acceleration of 1.70 m/s2. The car makes it one quarter of the way around the circle before skidding off the track. Determine the coefficient of static friction between car and track.

2. Three particles are connected by rigid rods of negligible mass lyuing around the y axis. (y=3, y=-2, y=-4). If the system rotates about the x-axis with an angular speed of 2 rad/s, find (a) the moment of inertia about the x-axis and the total rotationaly energy evaluated from 1/2Iw2 and (b) the linear speed of each particle and the total energy evaluated from (sigma)1/2mivi^2.

Just point me in the right direction...
 
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1.The friction force is a centripetal force.Equate the work done by the 2 forces & extract µ.
2.This is simple.What ideas do u have for it...?

Daniel.
 
forgot the weights...

y=3m, 4kg, y=-2m, 2kg, y=-4m, 3kg

I= (4)(3)^2 + (2)(2)^2 + (3)(4)^2
I= 92kg
 
O_o

RE= 1/2Iw^2

1/2(92)(2)^s

(46)(4)

184 J...
 
Yes,it's okay.Basically your problem's done. :smile: The last point is as easy as it was the first.

Daniel.
 

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