Question

Short Physics Question.

Answer 1

In the game of tether-ball, a 1.25 m rope connects a
M= 0.780 kg ball to the top of a vertical pole so that
the ball can spin around the pole as shown in the figure below.

http://i.imgur.com/Tb0DjOD.jpg

((Copy paste, don't click, might bring you the actual site,
which you don't really want...))

What is the speed of the ball as it rotates around the pole
when the angle θ of the rope is 36.0º with the vertical?

Answer 2

I don't really understand how to set this problem up, and how to identify the forces :(

Answer 3

I am thinking that since the ball is not accelerating vertically (or else it would fall or fly up), that the only forces I need to worry about are the horizontal forces.

I know that the radius in this case would be:
R=r•sinθ
R= (1.25 m) • (sin 36º) = 0.734732 m

And the centripetal force is:
F_c = m•v^2 / R

Is this correct so far?

Answer 4

a similar problem here :


Consider a conical pendulum with a 80 kg bob on a 10 m wire making an angle of θ = 5o with the vertical. Determine
(a) the horizontal and vertical component of the force exerted by the wire on the pendulum and
(b) the centripetal acceleration of the bob.

Solution:
(a) A free-body diagram of the bob is shown.

The bob does not change its vertical position, y = constant, vy = ay = 0. The vertical component of T must have magnitude mg.
Tcos(5o) = mg, T = (80 kg 9.8 m/s2)/cos(5o) = 787 N
The magnitude of horizontal component of T is Tsin(5o) = 68.6 N. The horizontal component of the force points towards the center of the circle.

(b) The horizontal component of T provides the centripetal (radial) acceleration ar.
Tsin(5o) = mar, ar = (68.6 N)/(80 kg) = 0.857 m/s2.
The speed of the bob is found from ar = v2/r, v = (arr)1/2.
Since r = (10 m)*sin(5o), we have v = 0.86 m/s.

Answer 5

are you sure that this is in fact a similar problem?