Uniform circular motion car problem

In summary, the maximum speed a cart can go without leaving the tracks on a circular arc is dependent solely on the square root of the radius and the acceleration due to gravity, regardless of the mass of the person or cart. This is because at the maximum speed, the normal force is zero, making the cart just about to lose contact with the ground. The equation to determine this speed is v = √(rg), where r is the radius and g is the acceleration due to gravity.
  • #1
S[e^x]=f(u)^n
23
0

Homework Statement


a cart with a person in passes over a circular arch at a constant speed v who's radius is r. Does the maximum speed the cart can go without leaving the tracks depend on the mass of the person, the mass of the cart, both, or neither?


Homework Equations


Fg=mg -Gravitational Force
Fc=(mv^2)/2r? -Centripetal force
Fn=? -Normal force


The Attempt at a Solution


Fc=Fg-Fn
i honestly don't know where to go from here? can anyone offer any hints... i need to figure it out by tomorrow afternoon. my gut says it depends on neither, but from math I've fiddled around with it seems to indicate it depends on both
 
Physics news on Phys.org
  • #2
S[e^x]=f(u)^n;1463608 said:

Homework Statement


a cart with a person in passes over a circular arch at a constant speed v who's radius is r. Does the maximum speed the cart can go without leaving the tracks depend on the mass of the person, the mass of the cart, both, or neither?


Homework Equations


Fg=mg -Gravitational Force
Fc=(mv^2)/2r? -Centripetal force
Fn=? -Normal force


The Attempt at a Solution


Fc=Fg-Fn
i honestly don't know where to go from here? can anyone offer any hints... i need to figure it out by tomorrow afternoon. my gut says it depends on neither, but from math I've fiddled around with it seems to indicate it depends on both
that "2" in the denominator of your Fc equation doesn't belong there. Your equation Fc = Fg -Fn is good.
What is the value of Fn just as the cart would leave the tracks?
 
  • #3
sorry about the 2, i don't know why i put it there. anyway Fn would be 0 on the cart, but so would the Fn for the person inside... wouldn't it? in which case the cart would leave the tracks at the same speed regardless of whether there was a person in there?... but is it still dependent on the weight of the car?
 
  • #4
Well, weight (mg) is pulling down while the centrifugal force (mv2/r) would cause the cart to go tangent to the track. For the cart to stay on the track those two forces must be equal.

Set the forces equal and see what happens with respect to mass.
 
  • #5
S[e^x]=f(u)^n;1463608 said:

Homework Statement


a cart with a person in passes over a circular arch at a constant speed v who's radius is r. Does the maximum speed the cart can go without leaving the tracks depend on the mass of the person, the mass of the cart, both, or neither?


Homework Equations


Fg=mg -Gravitational Force
Fc=(mv^2)/2r? -Centripetal force
Fn=? -Normal force


The Attempt at a Solution


Fc=Fg-Fn
i honestly don't know where to go from here? can anyone offer any hints... i need to figure it out by tomorrow afternoon. my gut says it depends on neither, but from math I've fiddled around with it seems to indicate it depends on both

what is special when the car is driven at the maximum speed so that it is just about to lose contact with the ground is that the normal force is zero. Isolate for the speed and you will have your answer. (btw, you should have mv^2/r, not mv^2/2r)
 
  • #6
nrqed said:
what is special when the car is driven at the maximum speed so that it is just about to lose contact with the ground is that the normal force is zero. Isolate for the speed and you will have your answer. (btw, you should have mv^2/r, not mv^2/2r)

thanks, from what i can figure out the velocity at which the cart(regardless of mass) leaves the tracks is dependent solely on the square root of the radius. which sounds about right to me
 
  • #7
S[e^x]=f(u)^n;1463947 said:
thanks, from what i can figure out the velocity at which the cart(regardless of mass) leaves the tracks is dependent solely on the square root of the radius. which sounds about right to me

correct. (and it depends on the acceleration due to gravity...the max speed on th emoon would be different ;-) )
 
  • #8
thanks!
 

FAQ: Uniform circular motion car problem

What is uniform circular motion?

Uniform circular motion refers to the motion of an object moving in a circular path at a constant speed. In this type of motion, the object's velocity is constantly changing, but its speed remains the same.

How is uniform circular motion different from non-uniform circular motion?

In uniform circular motion, the object's speed remains constant, while in non-uniform circular motion, the speed varies at different points along the circular path. This means that the object's velocity is changing in non-uniform circular motion.

How is centripetal force related to uniform circular motion?

In uniform circular motion, the object is constantly changing direction as it moves around the circular path. This change in direction requires a force to act on the object, known as centripetal force. This force is always directed towards the center of the circle and is what keeps the object moving in a circular path.

What is the role of velocity in uniform circular motion?

Velocity plays a crucial role in uniform circular motion as it determines the direction of the object's motion at any given point along the circular path. In uniform circular motion, the velocity vector is always tangent to the circle, meaning it is perpendicular to the radius of the circle at that point.

How does the radius of the circle affect the speed of the object in uniform circular motion?

The radius of the circle is directly related to the speed of the object in uniform circular motion. As the radius increases, the speed decreases, and vice versa. This is because the object has to cover a greater distance in the same amount of time as the radius increases, resulting in a decrease in speed.

Similar threads

Back
Top