Finding the normal at the bottom of a curved track

[sadpwner]

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?

Homework Equations

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.

 

[ehild]

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?

Homework Equations

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.

 

[sadpwner]

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?

 

[ehild]

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?

 

[sadpwner]

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

 

[ehild]

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?

 

[sadpwner]

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.

 

[haruspex]

N=W-Fc

This treats F_c 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.

 

[sadpwner]

This treats F_c 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?

 

[haruspex]

Then,

N=W+Fc?

If N is positive up and W is positive down, yes.

 

[ehild]

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.