A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long (a

Question

A 2.0 kg ball is attached to a vertical post with two strings, one 2.0 m long (attached at the top of the poll to the ball) and the other 1.0 m long (attached to the pole so that whenset in a circlethe 1.0m string makes a right angle withthe pole, this atachment point would be the cneter of the circle the ball makes). If the ball is set whirling in a horizontal circle, what is the minimum speed necessary for the lower string to be taught? If the ball has a constant speed of 6 ms-1, find the tension on each string.

Explanation / Answer

if both strings are straight, we know the angle the hypotenuse makes with the horizontal string will be cos (theta) = 1/2 or theta = 60 deg the vertical component of the tension in the upper string must support the weight of the ball (since the horizontal string can provide no vertical force), so that we have Tu sin 60 = 2 kg x 9.8m/s/s where Tu is the tension in the upper string Tu = 22.63N---> Answer for the horizontal string to have no tension in it, the upper string must provide all of the centripetal force, therefore we have that the horizontal component of Tu must equal m v^2/r: Tu cos 60 = m v^2/r where r = 1 m v = Sqrt[ r Tu cos 60/m] = Sqrt[1 x 22.63 cos 60/2] = 2.38 m/s ---> Answer T I = mv^2/R - mg /tan theta = (2*6^2/1) - (2*9.8/tan 60) = 72 -11.31 = 60.68 N ---> Answer

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