- #1

snormanlol

- 5

- 2

- 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.

-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.