Answer q11 in picture goes with the above figure

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Jun 23rd, 2015

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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

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Jun 23rd, 2015

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Jun 23rd, 2015

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Jun 23rd, 2015
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Jun 23rd, 2015
Oct 24th, 2017
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