How Is a Child's Swing Speed Calculated Using Energy Conservation?

[bphysics]

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 \theta₀.

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 \theta₀.

(b)

The tension in the rope at the lowest point is 1.5 times the weight of the child. Determine the value of cos \theta₀

(Known data)

• Initial Speed = 0 (v0 = 0)
• Rope length is R
• Rope mass is negligible (0)
• Initial angle from vertical is \theta₀.

Homework Equations

• \DeltaV + \DeltaK = (V_f - V_i) + (K_f - K_i)
• V_f + V_k = V_i + K_i
• 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 mgh_a in PE, where h_a = 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\theta₀.

a)

PE\theta = (M * g) * (R - Rcos\theta = PE at \theta).

(1/2)(1/2)(v⁰_max) = PE\theta.

(1/2)(1/2)(v⁰_max) = (M * g) * (R - Rcos\theta).

v² = (M * g) * (R - Rcos \theta ) / (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.

 

[learningphysics]

Hmmm... a little confused by the 1/2 's

this is right:

PE at theta = (M * g) * (R - Rcostheta)

so total energy at theta = (M * g) * (R - Rcostheta)

at the bottom potential energy = 0.

kinetic energy = (1/2)Mv^2

set: (1/2)Mv^2 = (M * g) * (R - Rcostheta)

solve for v...

for part b)... try to get v at the bottom using centripetal motion ideas... then you can solve for theta using the formula from part a).

 

[tnguyen628]

at theta do you mean the lowest point? because

"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 LaTeX graphic is being generated. Reload this page in a moment.0."

 

[tnguyen628]

as for part B it is easy after you get part a...

Fnet = ma

T - mg = ma

1.5mg - mg = ma
1.5mg - mg = (mv^2)/R

plug the answer you got in part a for v (it is (2Rg(1-costhetha))^1/2 <--- square root) then everything should cancle and you get costheta = .75

cheers!