Mechanics, Uniform Circular Motion

In summary: Can you tell me more about what you think the question is asking?In summary, the question is asking what the magnitude of the net force is when the car is moving from A to B.
  • #1
priscilla89
18
0

Homework Statement


A flat racetrack viewed from above has curves with radius of 50.0 meters and 100. meters. A car having a mass of 1.00 x 10^3 kilograms moves counterclockwise around the track at a constant speed of 20.0 meters per second. It takes the car 20.0 seconds to travel from C to D.

Radius = 50.0 meters and 100. meters
Mass = 1.00 x 10^3 kilograms
Speed = 20.0 m/s
Time = 20.0 s from C to D

1. What is the magnitude of the net force acting on the car while it is moving from A to B?
2. Calculate the net force acting on the car while it is moving from B to C.
3. Calculate the distance from C to D.
4. Compare the magnitude of the centripetal acceleration of the car while moving from D to A, to the magnitude of the centripetal acceleration of the car while moving from B to C.
5. Compare the magnitude of the car's momentum at D to the magnitude of the car's momentum at B.
6. Compare the magnitude of the centripetal acceleration of the car at A to the magnitude of the car's centripetal acceleration at A if additional passengers were riding in the car.

Homework Equations



2 = {Pie}r / t
ac = v^2 / r
Fc = ma{c}

The Attempt at a Solution


I'm sorry I just don't understand but I know the equations to use. Ok, but ill attempt to do it at least.
1. I don't know anything much about net force.
Would this be the equation? a = Fnet / m
I'm just really sorry but I only know about centripetal force, average speed and the centripetal acceleration. I don't know about the magnitude of the net force only magnitude of centripetal acceleration.

I would really appreciate it if you can help me. I do understand I should learn this but I wish I knew the equation to use for net force.

Okay I understand 91.
91. ac = v^2 / r
ac = (20)^2 / 50
ac = 8 for D to A

ac = (20)^2 / 100
ac = 4 for B to C

Is this correct? Thanks I really appreciate this.
 
Last edited:
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  • #2
Welcome to PF, Priscilla!
It is difficult to understand the question without seeing the diagram. Perhaps you could sketch it in a paint program, save the painting to a photo site like photobucket.com and post a link here.

Yes, Fnet or Ftotal is always equal to ma.
Likely the only force you have on those turns is Fc = mv²/r = ma so your calc is probably correct.
 
  • #3
priscilla89 said:

Homework Statement


A flat racetrack viewed from above has curves with radius of 50.0 meters and 100. meters. A car having a mass of 1.00 x 10^3 kilograms moves counterclockwise around the track at a constant speed of 20.0 meters per second. It takes the car 20.0 seconds to travel from C to D.

Radius = 50.0 meters and 100. meters
Mass = 1.00 x 10^3 kilograms
Speed = 20.0 m/s
Time = 20.0 s from C to D

1. What is the magnitude of the net force acting on the car while it is moving from A to B?
2. Calculate the net force acting on the car while it is moving from B to C.
3. Calculate the distance from C to D.
4. Compare the magnitude of the centripetal acceleration of the car while moving from D to A, to the magnitude of the centripetal acceleration of the car while moving from B to C.
5. Compare the magnitude of the car's momentum at D to the magnitude of the car's momentum at B.
6. Compare the magnitude of the centripetal acceleration of the car at A to the magnitude of the car's centripetal acceleration at A if additional passengers were riding in the car.

Homework Equations



2 = {Pie}r / t
ac = v^2 / r
Fc = ma{c}

The Attempt at a Solution


I'm sorry I just don't understand but I know the equations to use. Ok, but ill attempt to do it at least.
1. I don't know anything much about net force.
Would this be the equation? a = Fnet / m
I'm just really sorry but I only know about centripetal force, average speed and the centripetal acceleration. I don't know about the magnitude of the net force only magnitude of centripetal acceleration.

I would really appreciate it if you can help me. I do understand I should learn this but I wish I knew the equation to use for net force.

Okay I understand 91.
91. ac = v^2 / r
ac = (20)^2 / 50
ac = 8 for D to A

ac = (20)^2 / 100
ac = 4 for B to C

Is this correct? Thanks I really appreciate this.
Okay, here is a diagram for the problem. Thanks a lor!
 

Attachments

  • physics diagram.jpg
    physics diagram.jpg
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  • #4
Great, the diagram really helps. Your work so far looks very good.
I see some of the questions involve no calculations, like #1.
 

Related to Mechanics, Uniform Circular Motion

1. What is mechanics?

Mechanics is a branch of physics that deals with the study of motion and its causes, including forces and energy. It involves understanding how objects move, why they move, and what factors influence their motion.

2. What is uniform circular motion?

Uniform circular motion is a type of motion in which an object moves along a circular path at a constant speed. This means that the object covers equal distances in equal time intervals, moving in a circular motion without any change in its speed.

3. What is centripetal force?

Centripetal force is a force that acts towards the center of a circular path, keeping an object moving in a circular motion. It is the force that continuously changes the direction of an object's velocity, without affecting its speed.

4. What are some examples of uniform circular motion?

Some examples of uniform circular motion include the motion of planets around the sun, the motion of a Ferris wheel, and the motion of a car on a curved track.

5. How is uniform circular motion related to Newton's laws of motion?

Uniform circular motion is related to Newton's laws of motion in that it follows the first law of motion, which states that an object will continue to move in a straight line at a constant speed unless acted upon by an external force. In the case of uniform circular motion, the external force is the centripetal force that keeps the object moving in a circular path.

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