physics help on Center of mass, A 2.0 meter of rod of mass 5kg has two weights h

Question

physics help on Center of mass,

A 2.0 meter of rod of mass 5kg has two weights hanging on two sides of it. Weight A has mass of 1 kg and hands 0.2 m from the left end of the rod. Weight B has mass of 2 kg and hands 0.5 m from the right end of the rod. If I were to add another weight C with mass 3.0 kg, where should I place it under the rod so that it would balance the top of the triangle?

If I decided not to place the weight C under the rod, what is the angular momentum acting onto the rod?

Two question. Thank you. Would reward points once it is solved.

Explanation / Answer

torque on rod due to Weight A=1*10*.8=8Nm

torque on rod due to Weight B=-2*10*.5=-10Nm

so net torque=2Nm

to balance it let we put  weight C on xm length from left

so, 3*10*(1-x)=2

x=.9333m

angular acceleration=torque/I=2/10=.2rad/s^2 (I=moment of inertia = mr^2/2 for rod=10kgm^2)

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