A 2 00kg mass is attached to the end of a 1.50m string. The other end of the str

Question

A 2 00kg mass is attached to the end of a 1.50m string. The other end of the string is attached to the end of a vertical rod. The rod is rotated with an angular speed omega=3.00 radians/sec. First determine the angle theta between the rod and the string then determine the magnitude of the angular momentum of the mass relative to the attachment point of the string to the rod? A bullet of mass m is fired horizontally into a block of wood of mass M lying on table. The bullet remains in the block and they slide a distance d after the

Explanation / Answer

m = 2 kg
l = 1.5m
w = 3 rad/s

let tension in the string be T
By force balance

Tsin(theta) = mw^2r
Tcos(theta) = mg

so tan(theta) = w^2*r/g = w^2 * lsin(theta)/g = sin(theta)/cos(theta)
cos(theta) = g/w^2 *l = 9.8/9*1.5 = 0.7259
theta = 43.454 deg

angular momentum = Iw = mr^2 *w = m*l^2 sin^2 (theta) w = 6.385 kgm^2/s

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