Finding the Tension in a Rope (Circular motion)

[indecisive727]

An energetic father stands at the summit of a conical hill as he spins his 25kg child around on a 5.7kg cart with a 2.4m long rope. The sides of the hill are inclined at 22º. He keeps the rope parallel to the ground, and friction is negligible.

What rope tension will allow the cart to spin with the 17 ?

I tried working this problem backwards, from another example in my physics book but the answer came out wrong.
First, I converted 17 rpm to 1.780235837 rad/s.
Then, I plugged that value into omega=(v/r) and got 4.272566009 m/s.
After that I used the equation T=mv^(2)/r and came up with 233.5095763 N, which I rounded to 230 N.

Help?

 

[MostlyHarmless]

T does not equal mv^2/r.

T has x and y components. If you draw a free body diagram, you'll find that the x component of the tension must equal mv^2/r in order to keep the rope parallel with the hill.

 

[indecisive727]

So the answer I got is just tension in the x-direction? How do I find the tension in the y-direction?

 

[MostlyHarmless]

If you know Tx and the angle, you can go backwards from there.

How would you find Tx in any other situation?

 

[Astrum]

You need to split it into two components, x and y. You also need to pay attention to gravity.

 

[MostlyHarmless]

Oh yes, gravity. Drawing an accurate free body diagram will reveal all the secrets!

 

[haruspex]

It's a strange question. There's more information than necessary. You can work out the tension just from the masses and the rope angle.