So you're on a swing with a rope 5 m long, swinging out over a creek that's 3 m wide. At some point you are swinging all the way across the creek - at the top point you are right above the opposite side. How fast are you moving when you get back to the lowest point in your arc?? Thanks!
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If I get this vision correct, the fulcrum of the pendulum is centered over a 3 m wide creek and has a 5 m long rope.
The height of rise that the end of the rope makes is determined by the angle
the angle will have a hypotenuse of 5 and an opposite side of 1.5
a = inv sin(1.5/5) = 17.46 degrees
The vertical distance of rise will be the radius times (1 - cos(a))
or
h = 5 (1-cos 17.46)
h = 0.23 m
neglecting air resistance
as potential energy is mgh
and max velocity can be found from (1/2) mv^2
mgh = mv^2/2
2gh = v^2
v = (2gh)^0.5
v = (2(9.81)(0.23))^0.5
v = 2.13 m/s
The height of rise that the end of the rope makes is determined by the angle
the angle will have a hypotenuse of 5 and an opposite side of 1.5
a = inv sin(1.5/5) = 17.46 degrees
The vertical distance of rise will be the radius times (1 - cos(a))
or
h = 5 (1-cos 17.46)
h = 0.23 m
neglecting air resistance
as potential energy is mgh
and max velocity can be found from (1/2) mv^2
mgh = mv^2/2
2gh = v^2
v = (2gh)^0.5
v = (2(9.81)(0.23))^0.5
v = 2.13 m/s
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Height at top h = 1 m
PE=mgh=m*g*1
KE=0.5*m*v^2
0.5*m*v^2=m*g*1
v=sqrt(2g)=sqrt(2*9.81)=4.43 m/s
PE=mgh=m*g*1
KE=0.5*m*v^2
0.5*m*v^2=m*g*1
v=sqrt(2g)=sqrt(2*9.81)=4.43 m/s