A 0.7 kg mass is moving in a circle of radius 1.5 m on a flat frictionless table

Question

A 0.7 kg mass is moving in a circle of radius 1.5 m on a flat frictionless table at the end of a string. The speed of the mass is 3.8 m/s. The string routes through a hole in the center of the table and is held by you underneath the table. a) What is the angular momentum of the mass? (Give in kgm) b) What is the tension in the string? (give in Newtons) c) If you pull on the string so that the radius of the circle decreases to one half its former value, what is the new tension in the string?(Give in Newtons) d) How much work did you do to reduce the radius by a factor of one-half? ( Give in Joules)

Explanation / Answer

a) the angular momentum of the mass = Iw = mrv = 3.99 kgm b) the tension in the string =mv^2/r =6.7 N,,,,,,c) when radius is reduces angular momentum is conserved,,,,,,,,I1W1=I2W2,,,,,,,,,v2=7.6m/s,,,,,,,,new tension in the string = m(v2)^2/(r2)=53.9 N,,,,,,,,,,d) work = mv2^2/2 -mv1^2/2 =15.2 J

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