3) In the figure at right the ball with mass m, 10.0 kg is swinging around in a

Question

3) In the figure at right the ball with mass m, 10.0 kg is swinging around in a horizontal plane (magically supported so it does not fall downward!) as shown by the dashed line, and is connected to the mass m, 5.0 kg by a massless cord of length 10.0 m. The cord passes through a frictionless ring mounted on a support, shown by the heavy line. This allows the ball to swing in a circle of radius r without friction, and lets the cord slide freely through the ring (raising m, and nl mj increasing r. or lowering m, and decreasing ) a) Find the acceleration rate of m, as a function of r. Make the simplifying assumption that the ball's speed remains constant at all times and its path is always circular. i) If the ball's speed is 4.00 m/s and r is initially 3.00 m, find the initial acceleration of m, including its sign (before it has risen or fallen). ii) Will m, rise, stay where it is, or drop (changing r at the same time). b) Find the initial value of the tension T. c) Will m, continue to move in the same direction until going as high as possible (touching the ring) or as low as possible (10 m below it), or will it reach a stable equilibrium position between them? i) If m, reaches a stable position how far below the ring will it be?

Explanation / Answer

a)

motion m1 of the ball is given by

m1 a = m1g -T

T = mbv2/r , Tension in the string

acceleration of the mass m1    a = g - mb/m1 * v2/r

putting the values for

a = 9.8 - 10/5 * 16/3 = - 0.87 m/s

The acceleartion of the mass m1 is upward and it moves up

tension in the string T = 10 *16/3 = 53.33 N

c) as m1 moves up r increases

Tension T decreases

acceleration a = g - 32/r   , becomes 0 when

r = 32/g = 3.265 m

and from then the acceleration becomes +ve that is accelerates downwad.

when m1g   = mb v2/r , the system reaches a stable equilibrium

r= 3.265 m

length of the string l = 10 m

length below the ring = 10- 3.265 = 6.735 m

r = (10/5) * 16/9.8 =

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