A stick with a mass of 0.152kg and a length of 1.00m is pivoted about one end so

Question

A stick with a mass of 0.152kg and a length of 1.00m is pivoted about one end so it can rotate without friction about a horizontal axis. The meter stick is held in a horizontal position and released.

a)As it swings through the vertical, calculate the change in gravitational potential energy that has occurred.
Use 9.81m/s^2 for the acceleration due to gravity.

b)As it swings through the vertical, calculate the angular speed of the stick.

c)As it swings through the vertical, calculate the linear speed of the end of the stick opposite the axis.

d) Find the ratio of the speed of a particle that has fallen a distance of 1.00m, starting from rest, to the speed from part C.

Explanation / Answer

m = 152 kg; L = 1 m; g = 9.81 m/s2

(a) For a uniform body like the rod, the center of mass is at its geometric center, or at 0.5 m. At the bottom of its swing, the center of mass has dropped that distance, h = 0.5 m. So the change in gravitational potential energy is:

E = m g h = (152)(9.81)(-0.5) = -745.56 J

The answer is negative because the potential energy has decreased.

(b) The rotational kinetic energy increases by an amount equal to the decrease in gravitational potential energy (conservation of energy). The moment of inertia for a rod spinning about its end is I = (1/3)m L2

and rotational kinetic energy is E = (1/2)I 2 = (1/2)((1/3)m L2)2 = (1/6)m L2 2

E = (1/6)m L2 2 = (1/6)m (1)2 2 = (1/6)m 2

745.56 = (1/6)(152)2 = 25.332

2 = 745.56/25.33 = 29.43

= (29.43) = 5.43 rad/s

(c) v = r = L = (1)(5.43) = 5.43 m/s

(d) The ratio, R = (2gh)/v = (2(9.81)(1))/5.43 = (19.62)/5.43 = 0.816

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