Calculating Maximum Speed of a Child on a Swing

[pinkyjoshi65]

Energy transforations!

A student swings on a swing so that his centre of mass is located 2.2 m from the rest point where the rope is attached to the rail. If he swings so that his maximum amplitude causes the rope to make an angle of 49° with the vertical, calculate the child's maximum speed during the swing.


I don't know where to start!

 

[pinkyjoshi65]

ok...i've got a slight idea..we have to take the motion of the spring as a pendulum..

 

[learningphysics]

use conservation of energy. gravitational potential energy... kinetic energy.

 

[pinkyjoshi65]

ok..soo at the equlibrium position, we use total energy=mgh where h is 2.2m, and at the new postition the total energy is 0.5mv^2 yes..?

 

[learningphysics]

ok..soo at the equlibrium position, we use total energy=mgh where h is 2.2m, and at the new postition the total energy is 0.5mv^2 yes..?

almost there... 2.2m is not your h. 2.2m is the length of the rope (not exactly but it is the distance from the joint to the center of mass of the child)...

How high above the bottom does the child's center of mass reach? use 2.2m, 49 degrees and trig. that would be your h.

 

[pinkyjoshi65]

ok..so i use tan49= h/2.2, hence i can find h..but what abt the mass ??..and wait should'nt the energy at the equilibrium position be the kinetic energy, since h is 0..?

 

[pinkyjoshi65]

so when i take the total energy at the equilibrium position as the kinetic energy, and the total energy at the other position as the potential energy, and solve them, i got v as 7.04m/sec..is tht ok..?

 

[learningphysics]

ok..so i use tan49= h/2.2, hence i can find h..but what abt the mass ??..and wait should'nt the energy at the equilibrium position be the kinetic energy, since h is 0..?

no, it's not tan49 you need...

have a look at the first diagrams on this page:

http://dev.physicslab.org/Document.aspx?doctype=3&filename=OscillatoryMotion_PendulumEnergy.xml

see how they get height...

 

[pinkyjoshi65]

ok..soo by doing that i got v as 3.86m/s

 

[learningphysics]

ok..soo by doing that i got v as 3.86m/s

looks good. I get 3.851m/s though... it's just the rounding.

 

[cyprusad]

If $$l$$ is the length of the rope then, the height will be $$h = l(1 - \cos{\theta} )$$
The velocity of the boy is then simply $$\sqrt{2gh}$$ at the equilibrium position. At equilibrium position the velocity of a pendulum is maximum, so that's what you're looking for.
You get the equation by using simple conservation laws. (That of the potential energy at the height and the kinetic energy at the bottom.
I know you've got the answer, but I'm just reiterating the facts so that it becomes clear why you do what you do.
Moreover try to use variables rather than direct numerical values, you'll miss the physics involved, and it might get complicated unnecessarily.