(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

A child of mass M holds onto a rope and steps off a platform. Assume that the initial speed of the child is zero. The rope has length R and negligible mass. The initial angle of the rope with the vertical is [tex]\theta[/tex]_{0}.

http://img155.imageshack.us/img155/460/scan0001ok6.jpg [Broken]

(a)

Using the principle of conservation of energy, develop an expression for the speed of the child at the lowest point in the swing in terms of g, R, and cos [tex]\theta[/tex]_{0}.

(b)

The tension in the rope at the lowest point is 1.5 times the weight of the child. Determine the value of cos [tex]\theta[/tex]_{0}

(Known data)

- Initial Speed = 0 (v0 = 0)
- Rope length is R
- Rope mass is negligible (0)
- Initial angle from vertical is [tex]\theta[/tex]
_{0}.

2. Relevant equations

- [tex]\Delta[/tex]V + [tex]\Delta[/tex]K = (V
_{f}- V_{i}) + (K_{f}- K_{i})- V
_{f}+ V_{k}= V_{i}+ K_{i}- PE = mgh
- KE = (1/2)mv
^{2}= mgh- v
^{2}= 2gh- v
^{2}c = 2gha

3. The attempt at a solution

At point C, PE = 0

At point C, child has lost mgh_{a}in PE, where h_{a}= R (length of rope)

In PEs place, child has gained KE = (1/2)mv^{2}

Results in V^{2}= 2gha = 2 (g)(R) = V^{2}

Above does not acount for different angle from 90 degrees.

Therefore, we use h = R - Rcos[tex]\theta[/tex]_{0}.

a)

PE[tex]\theta[/tex] = (M * g) * (R - Rcos[tex]\theta[/tex] = PE at [tex]\theta[/tex]).

(1/2)(1/2)(v^{0}_{max}) = PE[tex]\theta[/tex].

(1/2)(1/2)(v^{0}_{max}) = (M * g) * (R - Rcos[tex]\theta[/tex]).

v^{2}= (M * g) * (R - Rcos [tex]\theta[/tex] ) / (1/2)(1/2)

Solve for v (cant write out correctly with LaTex)

So, does this look correct?

b)

I dont seem to understand how to solve this.

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# Homework Help: Conservation of Energy - Swing

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