Additional problem: a simple pendulum, consisting of a string (of negligible mass) of length l with a small mass m at the end, is initially held horizontal (θ = 90o) and then released. (a) what is the maximum velocity that the mass attains after release? (b) at what angle, θm, is the power delivered to the ball by gravity a maximum as the pendulum swings down? take θ = 0 when the pendulum is vertical.

a)   v = √ 2gL  abd  b)  θ = 45º

Explanation:

a) for this part we use the law of conservation of energy,

Highest starting point

       Em₀ = U = mg h

Final point. Lower

       Em₂ = ½ m v²

      Em₀ = Em₂

      m g h = ½ m v²

      v = √2g h

      v = √ 2gL

b) the definition of power is the relationship between work and time, but work is the product of force by displacement

     P = W / t = F. d / t = F. v

If we use Newton's second law, with one axis of the tangential reference system to the trajectory and the other perpendicular, in the direction of the rope, the only force we have to break down is the weight

     sin θ = Wt / W

     Wt = W sin θ

This force is parallel to the movement and also to the speed, whereby the scalar product is reduced to the ordinary product

     P = F v

The equation that describes the pendulum's motion is

    θ = θ₀ cos (wt)

Let's replace

    P = (W sin θ) θ₀ cos (wt)

    P = W θ₀ sint θ cos (wt)

We use the equation of rotational kinematics

    θ = wt

    P = Wθ₀ sin θ cos θ

Let's use

   sin 2θ = 2 sin θ cos θ

   P = Wθ₀/2 sin 2θ

This expression is maximum when the sine has a value of one (sin 2θ = 1), which occurs for 90º,

    2θ = 90

    θ = 45º

answer: he got 23

explanation:

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered