Consider a thin 24 m rod pivoted at one end. A uniform density spherical object
ID: 2203621 • Letter: C
Question
Consider a thin 24 m rod pivoted at one end. A uniform density spherical object (whose mass is 9 kg and radius is 4.9 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is I(rod) =(1/3)m L^2 and the moment of inertia of the sphere about its center of mass is I(sphere) =(2/5)m r^2. What is the angular acceleration of the rod immediately after it is released from its initial position of 41 degrees from the vertical? The acceleration of gravity g = 9.8 m/s^2. Answer in units of rad/s^2 My work {M(rod) * (.5L) + M(sphere) * (L+r)} / total mass = center of mass(CM) weight = rod + sphere = 9+9=18 t = weight*CM*sin @ {9(.5*24) + 9(24*4.9)} / 18 = 20.45 t = 18*20.45*sin 139 = 241.49 I = (1/3)L^2 + (2/5)r^2 +h^2 h = 24+4.9 = 28.9 I = 8+9.604+28.9 T/I=241.49/46.504=5.193=WRONG....where???Explanation / Answer
1)moment of inertia of rod along the pivot point is 1/3mr^2=233.3 and momet of intertia of sphere along same axis I = !cm+Mh^2 I = 2/5mr^2+M(R+l)^2 I = 484.3 so total MOI=233.33 + 484.3=717.6 2)5m 3)coserving energy we get... mg(l/2-l{(3)^1/2}/4)=717*w w=0.1 rad/sec