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Car travels with fwd force help!

  • Thread starter aliciaw0
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  • #1
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A 1900 kg car starts from rest and drives around a flat 40.0 m diameter circular track. The forward force provided by the car's drive wheels is a constant 1000N .

What are the magnitude of the car's acceleration at t= 10.0s ?

If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?

For part A, i used F=ma=mv^2/r but that answer was wrong so im not sure which equation to use now.

and for part b

u(rollling)= 0.02 so u(rolling)*Fn= Ffriction but im not sure how to find the time
 

Answers and Replies

  • #2
siddharth
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aliciaw0 said:
For part A, i used F=ma=mv^2/r but that answer was wrong so im not sure which equation to use now.
First of all, you have to identify the various forces acting on the car.

Notice that there are two components of the acceleration here. There is a tangential acceleration which is caused due to the car's engine and drive wheels and there is a centripetal acceleration as the direction of velocity is changing.

What is the force that provides the centripetal acceleration and allows the car to travel in a circle? This force must be friction which acts towards the center of the circular track.

The force in the tangential direction which allows the car to increase it's velocity is the force due to the car's engine and drive-shaft.

Now, can you draw the FBD of the car and frame the equations necessary to solve this problem?
 
  • #3
OlderDan
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aliciaw0 said:
A 1900 kg car starts from rest and drives around a flat 40.0 m diameter circular track. The forward force provided by the car's drive wheels is a constant 1000N .

What are the magnitude of the car's acceleration at t= 10.0s ?

If the car has rubber tires and the track is concrete, at what time does the car begin to slide out of the circle?

For part A, i used F=ma=mv^2/r but that answer was wrong so im not sure which equation to use now.

and for part b

u(rollling)= 0.02 so u(rolling)*Fn= Ffriction but im not sure how to find the time
For part A, there ar two components of acceleration. The first is the forward acceleration that is always in the direction of motiuon of the car, increasing its speed. This comes form the forward force. The second is the centripetal acceleration required to keep the car on the circular path. This force is provided by friction and is perpendicular to the direction of motion of the car. It has no effect on the speed, but its magnitude is related to the speed by the centripetal force equation.

For part B you are using the wrong coefficient. Rolling friction is not what keeps the car from sliding. You need the coefficient of static friction. For this part you need the contripetal component of the force from part A.
 
  • #4
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can someone let me know if im getting close to the right answer?

so far i have

forces in the y= Fn-Fg=ma=0 Fn=Fg

forces in the x= F(foward)-F(friction)=ma=mv^2/r
1000N-(u(static)*F(normal))=1900v^2/20

to finish the problem i solve for V and then divide by the change in time to get the acceleration.
 
  • #5
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aliciaw0 said:
forces in the x= F(foward)-F(friction)=ma=mv^2/r
1000N-(u(static)*F(normal))=1900v^2/20
How can this be ? You are saying that the forward force is along the same axis as the centripetal force, while they must be perpendicular.

Look, the centripetal force has NO influence on the magnitude of the velocity, only on the direction. So calculating the velocity after ten sec is really easy because you only need the given force devided by the object's mass. Just think of the object as moving on a linear path. Do you know what formula connects time, acceleration and velocity in this case ? Do not make it too difficult :wink:

Then, your equation for the vertical y axis is ok

Finally, the friction force opposes the centripetal force. When setting these two equal to zero you can get an value for v. From the first formula that you used to calculate the velocity after 10 sec, you will be able to calculate the time at which friction and centripetal force have the same magnitude

regards
marlon
 
  • #6
mukundpa
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Basically the external force acting on the car to accelerate it is friction force in tangential direction,(that's why in mud with very small friction it is difficult to accelerate the car, though the wheels are rotating) and the centripetal force (in radial direction) too is provided by friction. Now think is it possible that two seperate friction forces acting on the car simultaneously.
 
  • #7
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o man lol i made it SO much harder then it really was. lol well for anyone else struggling EVER with this problem like EVERYONE else said there are two components to the acceleration in this problem.

a= sqrt( a(radial) + a(tangetial) ) the tangential is given by the foward force. Which is solved by F=ma then with the acceleration you can find the radial one, by dividing by the time it takes to get the velocity.

after that you use v^2/r and then thats the centrip. accel then just do the pythag. theorm and you can find the magnitude of acceleration.

part b;

set the centrip accel (mv^2/r)=f(friction)=u(s)Fn then plug into the equation

Vf=Vi+a(tangential)(delta T)


yay haha THANKS everyone.
 
  • #8
mukundpa
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No, you are again forgetting total friction force ma(total) < u(s)Fn
 
  • #9
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i dunno thats exactly what i did and handed to my prof and my answers were right =] haha so i dunno at least its right? =] thanks though
 
  • #10
Doc Al
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aliciaw0 said:
i dunno thats exactly what i did and handed to my prof and my answers were right =] haha so i dunno at least its right? =] thanks though
Actually, mukundpa is correct. Realize that the static friction of the road on the tires provides the total horizontal force on the car, not just the centripetal component. The tangential component of the force is fixed (given as 1000N). The maximum force that friction can supply is given by [itex]\mu mg[/itex]. So the maximum centripetal force can be found by solving: [itex](\mu mg)^2 = 1000^2 + F_c^2[/itex]. Once you have the maximum centripetal force, you can find the maximum speed and then the time.
 

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