What is the centripetal force acting on a truck going around a circular path?

AI Thread Summary
A truck with a mass of 20,000 kg traveling at 50 km/hr on a circular path with a radius of 105 m requires the calculation of centripetal force. The correct formula to use is F = (m*v^2)/r, where speed must be converted to meters per second. The discussion clarifies that centripetal force (CPF) is not a standalone force but arises from the net force acting towards the center, typically due to friction. It emphasizes that centrifugal force is a fictitious force and should not be included in calculations from an inertial frame. Understanding the distinction between these forces is crucial for solving problems involving circular motion.
physics(L)10
Messages
101
Reaction score
0

Homework Statement


A truck with a mass of 20000kg goes around a circular path with a speed of 50km/hr. If the radius is 105m. Calculate the centripetal force acting on the car.


Homework Equations


This is why I can't get the answer to this question. I don't know which equation to use. I'm thinking its Fc=4pi^2mrf^2, but I don't know the frequency and speed isn't in the equation so it's probably not the right equation.


The Attempt at a Solution



Fc= 4pi^2(20000)(105)f^2
 
Physics news on Phys.org
See: http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html"
 
Last edited by a moderator:
So the answer would be F=(20000)(50^2/105)?
 
Oh wait, it says velocity, not speed. How would you findd that then?
 
speed refers to the magnitude of the velocity. SInce the problem requires you to find the MAGNITUDE of centripedal acceleration, the speed would be fine.
 
physics(L)10 said:
So the answer would be F=(20000)(50^2/105)?
Careful with units. Standard units for speed would be m/s, not km/hr.
 
Alright, thanks a lot for your help guys :)

Final answer:
F=(20000)([50x1000/60/60)^2]/105)
 
The centre of the circular path exerts this centripetal force, correct?
 
physics(L)10 said:
The centre of the circular path exerts this centripetal force, correct?

No. CPF itself doesn't have an existence. It comes into existence when a uniform circular motion takes place so as to balance the centrifugal force. It is always directed towards the centre though.
 
  • #10
physics(L)10 said:
The centre of the circular path exerts this centripetal force, correct?
No, it's the friction that exerts the centripetal force. That force points toward the center though.

RoughRoad said:
CPF itself doesn't have an existence.
Sure it does. (Assuming by CPF you mean centripetal force.)
It comes into existence when a uniform circular motion takes place so as to balance the centrifugal force.
There's no need to mention centrifugal force (which is a fictitious force), unless you are analyzing things from a non-inertial frame.
 
  • #11
Doc Al said:
No, it's the friction that exerts the centripetal force. That force points toward the center though.


Sure it does. (Assuming by CPF you mean centripetal force.)

There's no need to mention centrifugal force (which is a fictitious force), unless you are analyzing things from a non-inertial frame.

I know that CFF is a pseudo force but doesn't CPF arise because of this false force?
 
  • #12
RoughRoad said:
I know that CFF is a pseudo force but doesn't CPF arise because of this false force?
Pseudo forces only arise when applying Newton's laws in a non-inertial frame. Viewed from the usual inertial frame, centrifugal force never enters the picture. Something moving in a circle is centripetally accelerating; whatever force is causing that acceleration is labeled the centripetal force.
 
Back
Top