Mass on one side of a string

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1. In this lab, you will be rotating a mass on one side of a string that is balanced by a second mass on the other end of the string (Figure 6) "Figure 6 is attached". A free-body diagram will show that the centripetal force on the rotating mass m1, is provided by the weight of the hanging mass m2. Since those forces must be equal, we can write the equation: m2g = m1v^2/r where v is the velocity of m1, and r is the radius of its circular path. Since the magnitude of the velocity is the average distance divided by the average time, we can write the velocity = the circumference / the period, or v = 2πr/T where the period T is the time to complete one revolution. Assume m2 = 4m1. Write an expression for the period in terms of r and g. You should find the mass terms will drop out.

2. The around the world yo-yo trick is completed by twirling a yo-yo in a vertical circle. If the yo-yo was in uniform circular motion, compare the force of tension at the top of the circle to the force of tension at the bottom of the circle.

Hint: Drawing a free body diagram will be helpful.

Reference no: EM131461821

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