A4-lb ball b is traveling around in a circle of radius r1 = 3 ft with a speed (vb)1 = 6 ft/s. if the attached cord is pulled down through the hole with a constant speed vr = 2 ft/s, determine how much time is required for the ball to reach a speed of 12 ft/s. how far r2 is the ball from the hole when this occurs? neglect friction and the size of the ball.

t= 0.75 s

Explanation:

Here given that there is no any external torque that is why it angular momentum will be conserve.

Initial  angular momentum= m V₁ r₁

Final  angular momentum= m V₂ r₂

So

m V₁ r₁ =  m V₂ r₂

Now by putting the values

m V₁ r₁ =  m V₂ r₂

V₁ r₁ =   V₂ r₂

6 x 3 = 12 x r₂

r₂ = 1.5 ft

So the time t

t = r₂ / Vr

t= 1.5 / 2

t= 0.75 s

x+2 because it just doubles hon so the pattern can actually tell you the expression.

so for example,

if x=2 then 2+2 is 4

if x=4 then 4+4 is 8

and so on and so forth.

i really hope that this makes sense and you out in a way hon: )