Determine the coefficient of static friction between car and track

AI Thread Summary
To determine the coefficient of static friction between a car and a circular track, the friction force must be equated to the centripetal force required for the car's uniform acceleration. The car accelerates at 1.70 m/s² and skids off after traveling a quarter of the circle, allowing for the calculation of static friction. In a separate problem involving three particles rotating about the x-axis, the moment of inertia is calculated as 92 kg·m², leading to a rotational energy of 184 J. The linear speeds of the particles can be derived from their positions and the angular speed of 2 rad/s. The discussions highlight the importance of applying physics principles to solve these problems effectively.
HurricaneH
Messages
10
Reaction score
0
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...
 
Physics news on Phys.org
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.
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top