posted by Leon on .
A mass m1 = 6.0 kg sits on a frictionless table, and rotates at a constant speed
v = 3.0 m/sin a circle of radius R = 1.5 m. m1 is attached by a massless string which
passes through a hole in the centre of the table to a second m2 (which hangs freely
without moving). There is no friction present at the point where the string passes through
the hole in the table. The mass m2 is?
a)1.2kg b)5.2kg c)5.5kg d)2.8kg e)3.7kg
Since there is no motion of the vertical string up or down, the weight of m2 must balance the centripetal force of m1.
m1*V^2/R = m2*g
m2 = m1*V^2/(R*g) = 3.67 kg
That rounds off to answer (e)