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

In summary, the conversation discusses a child of mass M stepping off a platform while holding onto a rope with length R and negligible mass. The initial angle of the rope with the vertical is \theta0. Using the principle of conservation of energy, the conversation asks for an expression of the speed of the child at the lowest point in the swing in terms of g, R, and cos \theta0. After solving for v, the conversation moves on to determine the value of cos \theta0 given that the tension in the rope at the lowest point is 1.5 times the weight of the child. By using the formula for centripetal motion, the conversation arrives at the answer of cos \theta0 = 0.75.
  • #1
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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  • #2
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).
 
  • #3
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."
 
  • #4
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!
 

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

How does a swing demonstrate the conservation of energy?

A swing demonstrates the conservation of energy by converting potential energy into kinetic energy and vice versa. When a person sits on a swing at its highest point, they have the most potential energy. As they swing down, this potential energy is converted into kinetic energy, causing the swing to move faster. When the swing reaches its lowest point, all of the potential energy has been converted into kinetic energy. As the swing moves back up, the kinetic energy is gradually converted back into potential energy. This cycle continues as long as the swing is in motion, demonstrating the conservation of energy.

Is the conservation of energy always true for a swinging motion?

Yes, the conservation of energy is always true for a swinging motion. This is because energy cannot be created or destroyed, only converted from one form to another. In the case of a swing, potential energy is converted into kinetic energy and vice versa, but the total amount of energy remains the same.

Can friction affect the conservation of energy in a swinging motion?

Yes, friction can affect the conservation of energy in a swinging motion. Friction is a force that opposes motion and can cause energy to be lost in the form of heat. In the case of a swing, friction between the swing and the air or between the chains and the support structure can cause some energy to be lost, resulting in a decrease in the height of the swing over time.

How does the length of the swing affect the conservation of energy?

The length of the swing affects the conservation of energy by determining the speed and height of the swing. A longer swing will have a greater potential energy at its highest point, resulting in a higher swing. However, a longer swing will also have a longer period of oscillation, meaning it will take longer to complete one full swing. This affects the amount of kinetic energy that is converted back into potential energy and can result in a slower swing overall.

Is the conservation of energy applicable to all swinging objects?

Yes, the conservation of energy is applicable to all swinging objects, as long as there is no external force acting on the object. This means that the total amount of energy will remain the same, regardless of the size, shape, or material of the swinging object. However, factors such as air resistance and friction can affect the amount of energy that is converted and the overall motion of the swinging object.

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