How Much Force Keeps a Car on Track in Circular Motion?

[paperclip]

Hello all,
I also have a calculus thread in the appropriate forum- I wanted to have a separate physics thread in order to make things a little tidier around here!

A quickie:

A 1,200kg car turns a corner at 8ms^-1 and travels along the arc of a circle radius 9m in the process. How large a horiz. force must the road exert on the tyres in order to hold the car in a circular path & what is the min. coefficient of friction such that the car does not slide?

Find horizontal force=
Tangential force= mv^2/r
(1200kg)(8m/s)^2/9 m= 8533.33N

To keep from sliding off: the horizontal force and coefficient friction must work together against the centripetal force!

Centripetal force= m[PLAIN]http://theory.uwinnipeg.ca/physics/circ/img24.gifr.

m[PLAIN]http://theory.uwinnipeg.ca/physics/circ/img24.gifr, taking 8^2 as http://theory.uwinnipeg.ca/physics/circ/img24.gif

1200*64*9=691,200N

If the centripetal force is 691,200N, the opposing force must be able to deal with this: it is (where u=coefficient of friction) uF.

u*8533.33N=691,200N
u=81
too large.



Another go:

ug=[mv^2]/r



ug=[mv^2]/r, hence u (9.813)=[ (1,200kg)(8m/s^-1)]/9

=869.59

This seems like far too large of a coefficient of friction, so let's go about it by first principles.

u=Ff/FN. Ff=ma, a=rw^2

a=(9)(8)^2=576kg/ms^-1
m=1200kg, as above, so 1200kg *576kg/ms^-1= 691,200

Ff=691,200N
FN=Mg, 1200*9.813=11775.6

u=Ff/FN, so 691,200/111775.6= 6.183

This is still too large.. shouldn't coefficient of friction be LESS than one, otherwise it would slide right off the road??

The answer is lurking, but it's being a bit stubborn...help very greatly appreciated!

 

[Dick]

Centripetal force = mr!? Where did you get that? Go through this exercise and put units on everything. E.g. mr has units of kg*meter. That is NOT the units of a force.

 

[Dick]

And why is there a tangential force? And why does the expression that you wrote down for it involve a radial acceleration?

 

[malawi_glenn]

if have threads like this, it is much more complicated to see when you need help with a new thing ;)

 

[paperclip]

Dick, I put in a symbol there- mw^2r, not mr. It's possible that the symbol isn't showing up on your browser. Isn't friction plus tangential force the force opposing centripetal motion? As in, the tangential force is the horizontal force indicated in the question? I'm afraid the professor wasn't very clear on this point.

Malawi, the help that I get with this may help other students-putting a few questions together in one thread might help others categorise things more in their minds, plus if it's near the top and I'm the last person to post, chances are I'm asking a new question or getting new information.

 

[malawi_glenn]

It is more job for the others, the chances that you get help is lower when you do a thread like this ;)

 

[Dick]

Isn't friction plus tangential force the force opposing centripetal motion? As in, the tangential force is the horizontal force indicated in the question? I'm afraid the professor wasn't very clear on this point.

The professor must not have been. There aren't any mysterious concepts here that aren't present in linear motion. There is one force acting on the car (excepting the normal force/gravity pair) - friction. And that force causes a radial acceleration. It's that simple.

 

[husky88]

As Dick said, there are only 3 forces on the car. Fg = Fn and Fc.
Fg and Fn are opposite and they don't affect the centripetal acceleration. Maybe you are confused from when the road is inclined. If the road would be inclined then part of Fn would go in the calculation, as it would not be parallel to Fg anymore.
So Fc = Ff = m * v^2/r = 9000 N.
And coefficient of friction = Ff / mg = 0.7.

 

[paperclip]

Thanks so much for the help, guys! You're helping this strange language to settle itself into my brain.

Thanks for the pointers. Hmm, no new magical forces appear? I'll have to think about that one! No unicorns, then?

Wait a minute though...Fc=Ff=m*v^2/r?
so 1200x8^2/9? This gives me 8,533.33...
This however does give me the same coefficient of friction, so I guess it's OK?

Here's a bit of a toughie, for me at least:

Thunderstorms obtain their energy by condensing the water vapor in humid air. If a th understorm condenses all the water vapor in 10km^3 of air containing 1.74x 10^-2 kg m^-3 of water at 100% humidity, how much heat is released? The latent heat of vaporization of water is 2.256x10^6 J kg^-1.

Here's how I did it:
Q=mL
v=10km^3 air@ 100% humidity, density= 1.74x10^-2 kgm^-3

d=m/v, so 1.74x10^-2=m/10,000^3, m=174

Q=mL, Q= 174x 2.256x10^6, Q=3.925x10^8

Upon examination of the answer, the correct answer is 3.9x10^14. What the hey?

Thanks in advance!

 

[Dick]

You appear to be under the impression that 10,000^3=10^4. What should it be?

 

[paperclip]

1x10^12 which worked out gives a final answer of 3.9x10^16...closer, but still no cigar!

 

[husky88]

I said the answer is 9000 N because I rounded it off to two significant figures, but we got the same result. In some places, like where I live schools and universities consider the answer wrong if it is not with the proper number of significant figures. Meh...

 

[Dick]

1x10^12 which worked out gives a final answer of 3.9x10^16...closer, but still no cigar!

10(km^3)=10*(1000m)^3=10^10 m^3. Different from (10km)^3. This must be what they intended.