Consider a thin 36 m rod pivoted at one end A uniform densit

Consider a thin 36 m rod pivoted at one end. A uniform density spherical object (whose mass is 4 kg and radius is 3 m) is attached to the free end of the rod. The moment of inertia of the rod about an end is Irod = 13 mL^2 . The moment of inertia of the sphere about its center-of-mass is Isphere = 2 5 mr^2 . Note: The length Cmass in the figure represents the location of the center-of-mass of the rod plus mass system. The rod is initially at rest at 25 below the horizontal. The acceleration of gravity g = 9.8 m/s 2 . What is the angular acceleration of the rod immediately after it is released? Answer in units of rad/s 2 .

Solution

m = 4 kg, L = 36 m, r = 3 m, theta = 25 deg

I = I_rod + I_m = 1/3 ML² + m(L+r)² + 2/5 mr²

= 1/3 * 4 * 36² + 4(36+3)² + 2/5 * 4 * 3²

= 7826 kg/m²

Cmass= 3L/4 + r/2 = (3 * 36 / 4) + (3/2) = 28.5 m

alpha = T/L = c_mass (m+m) * g cos theta / I

= (28.5 * 8 * 9.8 * cos 25) /(7826)

= 0.259 rad/s²