Consider a thin 18 m rod pivoted at one end. A uniform density spherical object

Question

Consider a thin 18 m rod pivoted at one end. A uniform density spherical object (whose mass is 8 kg and radius is 3.1 m) is attached to the free end of the rod and the moment of inertia of the rod about an end is I(rod) = (mL^2)/3 and the moment of inertia of the sphere about its center of mass is I(sphere) = (2mr^2)/5. What is the angular acceleration of the rod immidiately after it is released from its initial position of 45 from the vertical? The acceleration of gravity is 9.8 m/s^2. Answer in units of rad/s^2.

Explanation / Answer

SOLUTION: Lenght of the thin rod, L=18 m Mass of the spherical object, m=8 kg Radius of the spherical object, r=3.1 m Moment of inertia of the Rod, Irod = mL2/3 Moment of inertia of the Sphere, Isphere = 2mr2/5 Angle made by the Rod of initial position from the vertical,=45o Moment of inertia about the pivot, I = Irod +[Isphere + m(L + r)2]    = mL2/3 + 2mr2/5 + m(L +r)2
Torque acting on the Rod, = mg(L/2)sin + mg(L + r)sin    =mg(3L/2 + r)sin
but, =I Here, Angular acceleration of the rod,        =/I          =(mg(3L/2 + r)sin)/(mL2/3 + 2mr2/5 + m(L +r)2)          =g(3L/2 +r)sin/[L2/3+ 2r2/5 + (L +r)2]          =(9.8 m/s2)(3x18 m/2 +3.1 m)(sin45o)/ [(18 m)2/3+2(3.1 m)2/5+(18 m+2.1 m)2]          =0.404 rad/s2          =(mg(3L/2 + r)sin)/(mL2/3 + 2mr2/5 + m(L +r)2)          =g(3L/2 +r)sin/[L2/3+ 2r2/5 + (L +r)2]          =(9.8 m/s2)(3x18 m/2 +3.1 m)(sin45o)/ [(18 m)2/3+2(3.1 m)2/5+(18 m+2.1 m)2]          =0.404 rad/s2

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