Finding the normal at the bottom of a curved track

In summary: The centripetal force is the resultant of the gravitational force (downward) and normal force (upward). It is better to write the equation for the forces as Fc =N-mg.

Homework Statement

A boy slides on a sled on the curved track sketched (black line). (Treat the boy and sled as a particle.) His mass plus that of the sled is m = 41 kg. Friction and air resistance are negligible (I want a sled like that!). His initial speed is v0 = 1.3 m.s–1. The height h = 4.3 m. At the bottom, the track has a radius of curvature (grey circle) R = 3.2 m. At the bottom of the track, what is the normal force exerted by the track on the sled?

ac=v^2/r
W=mg
Ui+Ki=Uf+Kf

The Attempt at a Solution

N+Fcentripetal=W

N=W-Fc

W=mg=-9.8*41

Ui+Ki=Uf+Kf

m*g*h+m*v0^2/2 =m*v^2

2*g*h+v0^2=v^2

2*-9.8*4.3+1.3^2=-82.59

I would plug this into the centripetal acceleration formula but v^2 shouldn't be a negative number so I am not quite sure if I got the right number.

Attachments

• cart track.png
2.9 KB · Views: 395

Homework Statement

A boy slides on a sled on the curved track sketched (black line). (Treat the boy and sled as a particle.) His mass plus that of the sled is m = 41 kg. Friction and air resistance are negligible (I want a sled like that!). His initial speed is v0 = 1.3 m.s–1. The height h = 4.3 m. At the bottom, the track has a radius of curvature (grey circle) R = 3.2 m. At the bottom of the track, what is the normal force exerted by the track on the sled?

ac=v^2/r
W=mg
Ui+Ki=Uf+Kf

The Attempt at a Solution

N+Fcentripetal=W

N=W-Fc

W=mg=-9.8*41

Ui+Ki=Uf+Kf

m*g*h+m*v0^2/2 =m*v^2

2*g*h+v0^2=v^2

2*-9.8*4.3+1.3^2=-82.59

I would plug this into the centripetal acceleration formula but v^2 shouldn't be a negative number so I am not quite sure if I got the right number.

g=9.8 m/s2, and the potential energy at height h is mgh >0 . It was incorrect to take it negative.

ehild said:
g=9.8 m/s2, and the potential energy at height h is mgh >0 . It was incorrect to take it negative.
Wow, I got it wrong somehow.

Continuing from where I left off:

2*9.8*4.3+1.3^2=85.97

41*9.8-41*85.97/3.2=-699.69N

The normal force is negative?

Wow, I got it wrong somehow.

Continuing from where I left off:

2*9.8*4.3+1.3^2=85.97

41*9.8-41*85.97/3.2=-699.69N

The normal force is negative?
What are these numbers? How do you get the normal force?

ehild said:
What are these numbers? How do you get the normal force?
N=W-Fc

=mg-mv^2/r

=41*9.8-41*85.97/3.2

=-699.69N

N=W-Fc

=mg-mv^2/r
That is not correct.
At what direction can the bottom of the track push the sled, upward and downward? What is the direction of the centripetal force? upward and downward?

ehild said:
That is not correct.
At what direction can the bottom of the track push the sled, upward and downward? What is the direction of the centripetal force? upward and downward?
Track pushes sled upwards. Centripetal is towards the centre so that is upwards as well.

N=W-Fc
This treats Fc as though it is an applied force, which it is not. It is that resultant of the applied forces necessary to achieve the curved path. Treat it much like the ma term in ΣF=ma.

haruspex said:
This treats Fc as though it is an applied force, which it is not. It is that resultant of the applied forces necessary to achieve the curved path. Treat it much like the ma term in ΣF=ma.
Then,

N=W+Fc?

Then,

N=W+Fc?
If N is positive up and W is positive down, yes.

Then,

N=W+Fc?
The centripetal force is the resultant of the gravitational force (downward) and normal force (upward). It is better to write the equation for the forces as Fc =N-mg.

1. What is the normal at the bottom of a curved track?

In physics, the normal at the bottom of a curved track refers to the perpendicular line that extends from the surface of the track at the point where the track curves. It is used to calculate the centripetal force acting on an object traveling along the curved track.

2. How is the normal at the bottom of a curved track calculated?

The normal at the bottom of a curved track can be calculated using the formula N = mg + (mv^2)/R, where N is the normal force, m is the mass of the object, g is the acceleration due to gravity, v is the velocity of the object, and R is the radius of the curved track.

3. Why is it important to find the normal at the bottom of a curved track?

Finding the normal at the bottom of a curved track is important because it helps us understand the forces acting on an object as it travels along the curved track. It is also necessary to ensure the safety and stability of the object and track.

4. How does the normal at the bottom of a curved track affect the motion of an object?

The normal at the bottom of a curved track affects the motion of an object by providing the necessary centripetal force to keep the object moving along the curved track in a circular path. It counteracts the centrifugal force acting on the object and prevents it from flying off the track.

5. Are there any real-life applications of finding the normal at the bottom of a curved track?

Yes, there are many real-life applications of finding the normal at the bottom of a curved track. For example, it is used in designing roller coasters, calculating the forces acting on cars as they navigate curves on a road, and determining the motion of satellites in orbit around a planet.

• Introductory Physics Homework Help
Replies
6
Views
1K
• Introductory Physics Homework Help
Replies
2
Views
931
• Introductory Physics Homework Help
Replies
7
Views
2K
• Introductory Physics Homework Help
Replies
5
Views
1K
• Introductory Physics Homework Help
Replies
1
Views
1K
• Introductory Physics Homework Help
Replies
7
Views
1K
• Introductory Physics Homework Help
Replies
4
Views
4K
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
6
Views
2K
• Introductory Physics Homework Help
Replies
11
Views
5K