# This is actually a college level question.

I'm having trouble with two of my Physics Problems. It goes a little something like this:
A bowling ball weighing w is attached to the ceiling by a rope of length L. The ball is pulled to one side and released; it then swings back and forth as a pendulum. As the rope swings through the vertical, the speed of the bowling ball is v.

What is the acceleration of the bowling ball, in magnitude and direction, at this instant? Take the free fall acceleration to be g.

And then the last:
A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investigate the motion in which the string moves in a cone with half-angle theta.

What tangential speed, v, must the bob have so that it moves in a horizontal circle with the string always making an angle theta from the vertical? Express the speed in terms of m, g, L, and/or theta.

I'm having a particularly hard time with the first one. THanks if you decide to help a brother out!!!

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nautica

Original Roational Velocity = Final rotational velocity + g * time

The rotational velocity will take into account the length

Nautica

joc
1. the question seems elementary...i'm not sure i'm reading it right.

the resultant acceleration of the ball is the vector sum of the tangential components and centripetal components of acceleration. however, since the ball is at the midpoint of the swing, there is a zero tangential component, because there are no forces acting on the ball tangentially to its path. the only 2 forces acting are the tension in the rope and the weight of the ball, both of which are perpendicular to its path.

therefore, the resultant acceleration of the ball is simply the centripetal component, which is directed towards the centre of rotation/swing (exactly upwards).

the circular motion equation applies here:

F = mv^2 / L
ma = mv^2 / L
a = v^2/L

that seems to be about it.

2. let A = theta. now obtain a general expression relating the required variables. first draw a diagram and resolve the forces into vertical and horizontal components:

(eqn 1) TsinA = mv^2 / r (using eqn of circular motion)

and

(eqn 2) TcosA = mg (since the ball's height in the air doesn't change)

r is the radius of rotation, and is equal to LsinA. T is the tension in the string.

dividing eqn (1) by (2),

tanA = v^2 / rg

substituting r = LsinA into eqn,

tan A = v^2 / gLsinA
v^2 = gLtanAsinA
v = sqrt [ gLtanAsinA ]

and that's it.

HallsofIvy