At what angle does a ball on a massless pendulum reach equilibrium?

[tomstringer]

Homework Statement

A rod of neligeable mass is released from the horizontal position. As a ball at the end of the rod falls, it reaches a point at which the tension, T, in the rod equals the ball's weight. At what angle from the verticle does this occur. I am not getting the same answer as my book--Halliday and Resnick 7th ed, problem 69, p 195.

Homework Equations

Let the verticle height the ball falls, h, h = R sinθ where R is the length of the rod.
Fnet on the ball = ma = T - Fg. T = mg(weight of ball). Fg = mgsinθ.
ma = mv^2/R(centripital force).
So if T = mg, then mg = mv^2/R + mgsinθ and hence sinθ = 1 - v^2/Rg.

The Attempt at a Solution

Forging on, 1/2mv^2 = mgh because the kinetic energy of the ball equals the gravitational work done on the ball at the point in question.
So v^2 = 2Rgsinθ.
Solving for θ, sinθ = 1 - 2Rgsinθ/Rg, 3sinθ = 1, θ = 19.47°
Thus the angle from the verticle is 45° + 19.5° = 64.5°.
The book gives ans answer of 71°

 

[Doc Al]

So v^2 = 2Rgsinθ.
Solving for θ, sinθ = 1 - 2Rgsinθ/Rg, 3sinθ = 1, θ = 19.47°
Thus the angle from the verticle is 45° + 19.5° = 64.5°.

Not sure why you are adding 45°. θ is the angle from the horizontal.

 

[tomstringer]

θ, the angle from the horizontal, was used to simplify the solution. Since the question asks, "what is the angle from the verticle" I added 45°.

 

[Doc Al]

θ, the angle from the horizontal, was used to simplify the solution. Since the question asks, "what is the angle from the verticle" I added 45°.

So you think the angle between vertical and horizontal is 45°?

 

[Doc Al]

Try this: If something makes an angle of 0° with the horizontal what angle does it make with the vertical?

 

[tomstringer]

Done. Thanks.