The bob rises to the same height on the opposite sde.
MgL(1 - costheta) + (1/2) M V^2
= M g L[1 -cos(thetamax)]
M cancels out.
At the bottom of the swing,
theta = 0 and
V^2 = 2 g L [1-cos(thetamax)] = Vmax^2
For half the velocity at bottom,
1 - cos(theta)max must be reduced by a factor of 4. Theta becomes 28 degrees.
suppose a simple pendulum was released from an angle of 58 degrees, with a length of 0.75m and mass of .15kg. what would be the speed of the bob at the bottom of the swing? To what height would the bob get to on the other side. What angle of release would give half the speed of that for 58 degree release angle at the bottom of the swing?
1 answer
1 answer