A thin rod with a length of 2 meters, mass of 250 g and a center of mass moment

Question

A thin rod with a length of 2 meters, mass of 250 g and a center of mass moment of inertia I = ML^2/12 is mounted on a fulcrum at a point 0.8 meters from the left end of the rod. At the left end of the rod is a point mass of 50 g that carries a charge of +0.5 mu C. At the right end is an 80 g point mass that carries a charge of -0.3 mu C. The rod is oriented perpendicular to a uniform electric field of 2780 N/C, as shown. Include gravitational effects in your responses and calculate the following: For the rod to remain in static equilibrium a third point charge is placed at point P, directly below the 80 g mass. What is the sign and magnitude of the charge? The electric field is suddenly reduced to zero. Calculate the initial angular acceleration of the rod.

Explanation / Answer

(a) For the rod to be in static equalibrium, we must have

(1) the total force on the beam is zero.

(2) the product of the perpendicular component of force with the distance to the pivot is same for each other.

d1 F1p = d2 F2p

Here, F1p is the perpendicular component of force F1 at distance d1 from pivot.

F2p is the perpendicular component of force F2 at distance d2 from pivot.

Using this concept, we can find required charge.

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