Tension and speed of bowling ball pendulum passing the equilibrium position

  • Thread starter snormanlol
  • Start date
  • #1
Homework Statement:
A bowlingball with mass m is hanging on a roof with a wire of length L. The ball is pushed out of equilibrium so that the ball makes an angle θ with the vertical. After letting go it makes a pendulum move. a) Determine the speed of the bowlingball when reaching the equilibrium position. b)Determine the tension of the wire in fuction of L,M and θ.
Relevant Equations:
mgh=1/2*m*v^2
F=m*a
For part a I used conservation of energy.
-m*g*cos(θ)*L+1/2*m*0^2=-m*g*L +1/2*m*v^2 => v = sqrt(2*g*L(1-cos(θ )).
b) For b I was think that T = mg in the equilibrium point but that doesn't invole θ in the answer. So that's why I tought that T*cos(θ ) = mg. So that the tension is mg/cos(θ). But this isn't correct. The answer has to be T = mg*(3-2*cos(theta)).
Thanks for the help in advance. And I apologize for my bad english it isn't my native language.
And here is a homemade sketch of the problem.
 

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Answers and Replies

  • #2
BvU
Science Advisor
Homework Helper
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Hello snorman,
think that T = mg in the equilibrium point
That would be correct if the ball were not moving. But it is, and the kind of trajectory it describes should help you think of what else ##T## has to do except keeping the ball from accelerating downwards :smile:
 
  • #3
Hi bVU,
Thanks for the tip. arad= v^2/L. So that a = 2*g*(1-cos(theta)).
T - mg = 2*m*g*(1-cos(theta)) => T =mg(3-2*cos(theta))
 

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