Question

Two lady bugs of mass m land at the opposite ends of a stick of mass ml and length 1 that is supported at a point P located 1/51 from one end. The stick starts to rotate. Find the net torque and rotational inertia at the moment when the stick makes an angle a with respect to the horizontal.

Answer 1

Vao- A and B are two Aas ends of the rod. - |19-awa C is the center of mass of the rod. The two bugs applies force mg on each end of the rod. The gravitational force on the rod can be modeled by the force mag acting at the Center of the rod. All the forces makes angle 90-a with the rod. Distances AP = f = 0.21 PC = f 40.5 -0.21=0.30 P8 - PC+CB = 0.38 +0.50 = 0.8 l Net torque on the rod about point P is T=TA + Te + TB I - Fa PA) sin(90-«) – (PC) sinq0-2) – Fe (PB) sin 10-)

Force at A tends to rotate the rod in the counterclockwise direction, so it is taken as positive. While the forces at C and B tends to rotate the rod in the clockwise direction, so they are taken as negative.

T = mg (0.21) cosa- mag (0.31) cosa - mg (0.8 l) cosa 7 --(0.6m +0.3 m.) gf cosa Rotational inertia of the rod about point C is Inod na m 2 Using parallel axis theorem, the moment of inertia of the rod about point P is I noap = I had + me (PC)2 Inod p = 1 m 2 + me (0.3 0² Irsap = 13 m, 1² Moment of inential of the bug at A in IA = m (AP?? _ mx (4) - me Moment of inentia of the bug at B is Io = M (PB) = m 142133 16 m€ 25

Total moment of inestia about point P is Ip = Inoap + In + To Ir = 3 mil + mga + lama Tp = (13 m2 + 17m) en