what is the maximum velocity of a 6 kg object oscillation on a 4 m string. If the mass starts when the string makes a 45 degree angle with horizontal when the object is released from rest
physics - drwls, Friday, May 27, 2011 at 7:49am
Use energy conservation.
Potential energy change = maximum kinetic energy
M g L (1 - cos 45) = (1/2) M Vmax^2
Mass M cancels out. L is the string length
Vmax^2 = g*L*(0.5858)
Vmax = 0.7654 sqrt(g L)