One end of a string is attached to a center pivot, the other end to a 175 gram o

Question

One end of a string is attached to a center pivot, the other end to a 175 gram object. The

distance from the pivot to the center of mass of the object is 12 cm. The mass swings around the pivot 100

times each 60 seconds at constant speed. speed of mass=1.26m/s and tension of string =2.3N

NOW,

The same string hung vertically (no rotation), supporting the 175 gram

object against gravity (in equilibrium).

a) What is the tension in the string?

b) How much mass would have to be added to the 175 gram object to make the tension the same as

in problem 1(b)?

Explanation / Answer

a)

Cercumference = Pi * d = 3.14... * 0.24 m = 0.75 m

100 rev/60 s = 1.67 rev/s * 0.75 m = 1.26 m/s Tangential Velocity

b) I like to draw the Free Body Diagram with Centrifugal Force. I think that is more intuitive and easier for me to explain. Your teacher may not agree.

Cf is OUT and is the Tension in the string

Cf = Mass * tangential velocity ^2 / r = 0.175 kg * 1.26^2 / 0.12 m = 2.30 N

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.