A thin uniform rod (mass 4.4 kg, length 3.2 m) rotates freely about a horizontal

Question

A thin uniform rod (mass 4.4 kg, length 3.2 m) rotates freely about a horizontal axis A that is perpendicular to the rod and passes through a point at a distance d = 0.70 m from the end of the rod. The kinetic energy of the rod as it passes through the vertical position is 24 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

Given,

m = 4.4 kg ; L = 3.2 m; d = 0.7 m

a)The moment of inertia will be:

I = m l^2/12 + m (l/2 - d)^2

I = 4.4 x 3.2^2/12 + 4.4 (3.2/2 - 0.7)^2 = 7.32 kg-m^2

Hence, I = 7.32 kg-m^2

b)KE = 1/2 I w^2

w = sqrt(2 KE/I)

w = sqrt (2 x 24/7.32) = 2.56 rad/s

Hence, w = 2.56 rad/s

c)KE = m g (L/2 - d)(1 - cos(theta))

1 - cos(theta) = 24/4.4 x 9.8 x (3.2/2 - 0.7) = 0.618

cos(theta) = 1 - 0.618 = 0.382

theta = cos^-1(0.382) = 67.54 deg

Hence, theta = 67.54 deg

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