The diagram shows a thin rod of uniform mass distribution pivoted about one end

Question

The diagram shows a thin rod of uniform mass distribution pivoted about one end by a pin passing through that point. The mass of the rod is 0.450 kg and its length is 2.00 m. When the rod is released from its horizontal position, it swings down to the vertical position as shown. A thin rod labeled M is initially horizontal, with a pivot on its left end. The rod then rotates clockwise by its left end until it is hanging down vertically. There is a point marked CG a distance L/2 down the rod. (a) Determine the speed of its center of gravity at its lowest position. Incorrect: Your answer is incorrect. . Consider the conservation of energy of the center of mass of the rod. What is the relationship between angular and tangential velocities? m/s (b) When the rod reaches the vertical position, calculate the tangential speed of the free end of the rod. Incorrect: Your answer is incorrect. . Check your calculations. m/s

Explanation / Answer

a) From conservation of energy

m g * [L/2] = 1/2 I w2

m g * [L/2] = 1/2 (m L2 / 3) w2

3 g = L w2

w= sqrt [3 g / L]

= sqrt [3 * 9.8 / 2.00] = 3.83 rad/s

v = 3.83 * L/2

speed of its center of gravity v = 3.83 m/s

b)  tangential speed of the free end of the rod = 3.83 * L

= 3.83 * 2.00

= 7.66 m/s

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