##### Answer q11 in picture goes with the above figure

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Jun 23rd, 2015

There are two forces acting on the ball: gravity and tension in the string. These are in balance while twirling becuase the ball stays in the horizontal plane and at a fixed distance r = L sin(28) from the axis of rotation. L = 2.2 m.

Let T = tension on the string. The w = T cos(28) is the vertical component and h = T sin(28) is the horizontal component.

When the net forces are zero (in balance), we have W = mg = T cos(28) = w and C = mv^2/r = T sin(28) = h; where W and C are the weight and centrifugal force of the twirling ball.

a) The tangential velocity (speed) is found from v^2 = Tr sin(28)/m = TL sin(28)^2/m and T = mg/cos(28); so that v^2 = (mg/cos(28))L sin(28)^2/m = gL tan(28)sin(28) and v = sqrt(gL tan(28) sin(28)) for the tangential velocity (speed). You can do the math.

b) As tangential velocity v = wr; then w = v/r = v/(L sin(28)) is the angular velocity in rad/sec. So to travel once around C = 2 pi rads = wT; the period T = 2 pi/w = ? you can do the math.

Notice the mass m falls out and is not a factor. This results because weight and centrifugal force depend on mass and they balance out; so the masses cancel.

Please let me know if you nea case for torture by michale levined any clarification. I'm always happy to answer your questions.
Jun 23rd, 2015

Not understanding

Jun 23rd, 2015

Not understanding

Jun 23rd, 2015

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Jun 23rd, 2015
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Jun 23rd, 2015
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