In the figure, a thin uniform rod (mass 4.0 kg, length 4.0 m) rotates freely abo
ID: 2002193 • Letter: I
Question
In the figure, a thin uniform rod (mass 4.0 kg, length 4.0 m) rotates freely about a horizontal axis A that is perpendicular to the rod and passes through a point at a distance d = 1.1 m from the end of the rod. The kinetic energy of the rod as it passes through the vertical position is 19 J. (a) What is the rotational inertia of the rod about axis A? (b) What is the (linear) speed of the end B of the rod as the rod passes through the vertical position? (c) At what angle will the rod momentarily stop in its upward swing?
Explanation / Answer
kinetic energy of rod in vertical position = gravitational potential energy of rod at highest point in swing
19 J = mgy
The center-of-mass of the rod is initially at a height of half the length of the rod (4.0 m / 2 = 2.0 m) or 0.9 m (= x) below the axis A.
(a) rotational inertia = mL2/12 + m(4/2 - 1.1)2 = (4 x 16/12) + (4 x 0.81) = 8.573 kgm2
(b) VB = w(4-1.1) = sqrt(2 x 19/8.573) x 2.9 = 6.105 m/s
(c) At maximum the center-of-mass would be a distance of x*cos below axis A, which makes y = x - x*cos.
19 J = mgx(1 - cos)
19 J = (4.0 kg)(9.8 m/s²)(0.9 m)(1 - cos)
cos 0.46145
62.52°