Question

A mass of 0.875 kg, on a frictionless table, is attached to a string which passes through a hole in the table. The hole is at the center of a horizontal circle in which the mass moves with constant speed. The radius of the circle is 0.520 m and the speed of the mass is 17.0 m/s. It is found that drawing the string down through the hole and reducing the radius of the circle to 0.370 m has the effect of multiplying the original tension in the string by 4.63. Compute the total work (in joules) done by the string on the revolving mass during the reduction of the radius.

Answer 1

here angular momentum is conserved

angular momentum = r x p

p = mv  

L = mvr

r1*m*v1 = r2*m*v2

v2 = r1*v1/r2

given :

r1 = 0.520 m

v1 = 17 m/s

r2 = 0.370 m

v2 = 23.89 m/s

work = change in kinetic energy

w = KE2 - KE1

w = mv2^2/2 - mv1^2/2

w = 123.25 J