A 235 gram ball is attached at the right end of a uniform density bar. The bar m

Question

A 235 gram ball is attached at the right end of a uniform density bar. The bar mass is 455 grams. The bar is attached to a pivot near its left end. As shown in the diagram, a string on the bar's right end keeps the bar suspended at an angle of 40.0 degree . Throughout this problem you can assume that the radius of the ball is negligible compared to the length of the bar. Find the horizontal and vertical components of the force exerted by the pivot onto the bar. You may leave your answer in component form. The string is cut. Find the magnitude of the angular acceleration of the bar immediately after the string is cut. You may assume the bar is uniform

Explanation / Answer

Counter-clockwise torque = (L/4)/2 * .455 / 4 * g cos 40 + T (3 L/4) Clockwise torque = [(3 L/4)/2 * (3 * .455) /4 + .235 * (3 L/4)] * g * cos 40 Equate and solve for T where T is the tension in the string Horizontal force = T cos 50 Vertical force = (.235 + .455) g - T sin 50 Torque when string is cut = T * 3 L/4 (torque that maintained equilibrium) I = 1/3 * [(L/4)^2 * (.455/4) + (3 L/4)^2 * (3 *.455) /4] + .235 * (3 L/4)^2 since I for a uniform bar about 1 end is 1/3 M L^2 angular acceleration = Torque / I It appears that L will appear in the value for the angular acceleration

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