Centrepetal Problem help PLEASE

  • Thread starter jigs90
  • Start date
In summary, the conversation discusses two problems involving a car of mass 474 kg traveling around a circular race track and a hill with different radii. The maximum speed v that the car can go without flying off the track is determined using the equation for centripetal force, where static friction provides the force. The second problem also involves finding the maximum speed, but the force providing the centripetal force is different. The conversation also touches on using the correct values and units in calculations.
  • #1
jigs90
19
0
Centrepetal Problem help PLEASE!

Homework Statement


I don't even know where to start on either of these two problems

A car of mass 474 kg travels around a flat,
circular race track of radius 197 m. The co-
efficient of static friction between the wheels
and the track is 0.123.
The acceleration of gravity is 9:8 m=s2 :
What is the maximum speed v that the car
can go without flying off the track?

It has a follow up too which I would really appreciate some help on!

The same car now travels on a straight track
and goes over a hill with radius 101 m at the
top.
What is the maximum speed that the car
can go over the hill without leaving the road?


A step by step outline would make my day!
Thanks


Homework Equations



A= V^2/ r
F= MV^2/ r


The Attempt at a Solution



I'm not sure how to go about tackling that problem, I tried just substituting 101 m in for the radius but that doesn't work.
 
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  • #2
I tried doing the f= mv^2/r and when I entered in my answer for the first problem, it was wrong, I don't know how to take into consideration the static friction because I then used the equation mus (static friction) times N and subtracted that from the weight and then reworked the problem but my answer was still incorrect
 
  • #3
jigs90 said:
A car of mass 474 kg travels around a flat,
circular race track of radius 197 m. The co-
efficient of static friction between the wheels
and the track is 0.123.
The acceleration of gravity is 9:8 m=s2 :
What is the maximum speed v that the car
can go without flying off the track?

Both problems will involve the centripetal force equation, but centripetal force is not a physical force, so in each situation, you need to think about what force provides the centripetal force.

For this first one, the car is on a horizontal curve. What force is acting on the tire surfaces to hold it on the turn? How do you calculate that force? That is the force which provides the centripetal force, so you can set it equal to F_centripetal.


The same car now travels on a straight track
and goes over a hill with radius 101 m at the
top.
What is the maximum speed that the car
can go over the hill without leaving the road?

In this problem, the centripetal force is toward the center of the hill, which we are to take as having a semicircular cross section. At the top of the hill, what forces are acting on the car? The centripetal force will be equal to the net force on the car. What is true about these forces when the car is on the verge of losing contact with the road surface when it is moving fast enough?
 
  • #4
on the first one I set the equation of
mus(static)N = mv^2/ r but that still isn't right...Isn't the static friction providing the centripital force?
 
  • #5
jigs90 said:
on the first one I set the equation of
mus(static)N = mv^2/ r but that still isn't right...Isn't the static friction providing the centripital force?

It does sound weird to use static friction when talking about a moving car, but consider that, in order to hold the turn, we do not want the tires to slide along the road surface. (Likewise, when we deal with rolling wheels, we will also use static friction rather than kinetic friction.)

So, yes, the static friction is providing the centripetal force for a level road curve. How do you find N?
 
  • #6
isn't N equal to mg?
 
  • #7
jigs90 said:
isn't N equal to mg?

Yes, for an object on a horizontal surface. What do you get when you put these pieces together?
 
  • #8
isn't that what I had been using before though?
mus(static) mg= mv^2/s and the m's cancel and you solve for v^2 but I still keep getting it incorrect
 
  • #9
jigs90 said:
isn't that what I had been using before though?
mus(static) mg= mv^2/s and the m's cancel and you solve for v^2 but I still keep getting it incorrect

What are the values you're putting into calculate v^2?
 
  • #10
.123(474)9.8
 
  • #11
jigs90 said:
.123(474)9.8

What is the radius of the curve?
 
  • #12
Nevermind...I figured it out. I was entering in the wrong decimal place. Thanks for all your help though. :)
 

1. What is the Centrepetal Problem?

The Centrepetal Problem is a physics problem that involves finding the acceleration of an object moving in a circular path. It is also known as the Centripetal Force Problem.

2. What are the key concepts involved in the Centrepetal Problem?

The key concepts involved in the Centrepetal Problem include centripetal force, centripetal acceleration, and tangential velocity. These concepts are used to understand the motion of objects moving in a circular path.

3. How do you solve the Centrepetal Problem?

To solve the Centrepetal Problem, you need to use the formula a = v^2/r, where a is the centripetal acceleration, v is the tangential velocity, and r is the radius of the circular path. You also need to consider the direction of the acceleration, which is always towards the center of the circle.

4. What are some real-life applications of the Centrepetal Problem?

The Centrepetal Problem has many real-life applications, such as understanding the motion of objects in circular motion, such as a car going around a curve or a satellite orbiting the Earth. It is also used in designing roller coasters and other amusement park rides.

5. What are some common mistakes made when solving the Centrepetal Problem?

Some common mistakes when solving the Centrepetal Problem include forgetting to consider the direction of the acceleration, using the wrong formula, and using incorrect values for velocity or radius. It is important to carefully read the problem and double-check all calculations to avoid these mistakes.

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