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bphysics
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Homework Statement
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
(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.
Homework Equations
- [tex]\Delta[/tex]V + [tex]\Delta[/tex]K = (Vf - Vi) + (Kf - Ki)
- Vf + Vk = Vi + Ki
- PE = mgh
- KE = (1/2)mv2 = mgh
- v2 = 2gh
- v2c = 2gha
The Attempt at a Solution
At point C, PE = 0
At point C, child has lost mgha in PE, where ha = R (length of rope)
In PEs place, child has gained KE = (1/2)mv2
Results in V2 = 2gha = 2 (g)(R) = V2
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)(v0max) = PE[tex]\theta[/tex].
(1/2)(1/2)(v0max) = (M * g) * (R - Rcos[tex]\theta[/tex]).
v2 = (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 don't seem to understand how to solve this.
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