Uniform Circular Motion, Find the Tension

[Hoophy]

IMPORTANT!
TEXT IN GREEN HAS BEEN ADDED AND IS CORRECT
TEXT IN RED HAS BEEN REMOVED AND IS INCORRECT

1. Homework Statement
Question: An energetic father stands at the summit of a conical hill as he spins his 25 kg child around on a 5.7 kg cart with a 2.3-m-long rope.
The sides of the hill are inclined at 22∘.
He keeps the rope parallel to the ground, and friction is negligible.
What rope tension will allow the cart to spin with the 16 rpm

Homework Equations

My variables and constants:
Child Mass = m_child = 25 kg
Cart Mass = m_cart = 5.7 kg
Rope Length = L = 2.3 m
Radius = r = Lcos(22∘) = 2.3 m
Hill Grade = θ = 22∘
Angular Velocity = ω = 16 rpm = 1.67551608 rad/s
Tension = T
Total Mass = m = m_child + m_cart
Centripetal Acceleration = a_c
g = 9.8 m/s²
Weight = w = mg

The Attempt at a Solution

a) I drew a free body diagram with
1) Tension pointing North-West (up the slope and parallel to the surface)
2) Normal Force pointing North-East (perpendicular to the surface)
3) Weight pointing down
b) I defined my coordinate system as:
1) Positive x is up the slope and parallel to the surface
2) Positive y is parallel with the Normal Force and perpendicular to the surface (pointing away from the ground, into the sky)

c) I summed the forces in the x axis:
F_x: ma_c*cos(θ) = T - mg*sin(θ)
d) I isolated Tension:
T = ma_c*cos(θ) + mg*sin(θ)
e) I found a_c:
a_c = ω²r
f) I substituted a_c into my equation from (d):
T = mω²r*cos(θ) + mg*sin(θ)
!) T = mω²(L*cos(θ))*cos(θ) + mg*sin(θ)
g) I plugged in known values:
T = (25 kg + 5.7 kg)(1.67551608 rad/s)²(2.3 m) + (25 kg + 5.7 kg)(9.8 m/s²)*sin(22∘)
T = (25 kg + 5.7 kg)(1.67551608 rad/s)²((2.3 m)*cos(22∘))*cos(22∘) + (25 kg + 5.7 kg)(9.8 m/s²)*sin(22∘)
h) I computed Tension as T = 310.928 N = 310 N
T = 283.111 N = 280 N

4. A note
310 N is not the correct answer, nor is 310.928 N. Also I do not know what the correct answer is...
I was wondering if someone could help me figure out what it is that I am doing incorrectly?
I really appreciate the help, thanks in advance!
Thank you so much for helping me with this problem @TSny!

 

[TSny]

It would be helpful if you could post a diagram.

c) I summed the forces in the x axis:
F_x: ma_c = T - mg*sin(θ)

Does the cart move on a horizontal circle? If so, would the centripetal acceleration be parallel to your x-axis which is along the slope?

 

[Hoophy]

It would be helpful if you could post a diagram.Does the cart move on a horizontal circle? If so, would the centripetal acceleration be parallel to your x-axis which is along the slope?

I have edited my post to include a drawing.

I see...
So would I need to do:
Fx: ma_ccos(θ) = T - mg*sin(θ) ?
Thanks.

 

[TSny]

I have edited my post to include a drawing.

I see...
So would I need to do:
Fx: ma_ccos(θ) = T - mg*sin(θ) ?
Thanks.

Yes. Looks good.

Does r equal the length of the rope?

 

[Hoophy]

Yes. Looks good.

Does r equal the length of the rope?

Would I be able to replace all previous instances of "r" with (Rope_Length)cos(θ) ?

 

[TSny]

Would I be able to replace all previous instances of "r" with (Rope_Length)cos(θ) ?

Sounds right

 

[TheMercury79]

It may help if you consider the components of the tension. All of the tension in the rope does not account for the centripetal acceleration.
One component of the tension counters the weight, which in turn could affect the normal force. What new set of equations would you end up with if you introduce components of the tension?

 

[Hoophy]

Sounds right

That did the trick, thank you so much for all of your help!

I have amended my post with the correct process in case anyone with a similar question stumbles upon this thread.

 

[TSny]

That did the trick, thank you so much for all of your help!

I have amended my post with the correct process in case anyone with a similar question stumbles upon this thread.

Good work