Please show steps if possible. 012 (part 1 of 3) 7.0 points Consider a uniform r

Question

Please show steps if possible.

012 (part 1 of 3) 7.0 points Consider a uniform rod with a mass m and length L pivoted on a frictionless horizontal bearing at a point O (4/5 L form the lower end), as shown. What is the moment of inertia of the rod about the point O? The moment of inertia of a uniform rod about its center of mass is f/12 M L2. I = 13/147 M L2 I = 13/81 M L2 I = 1/9 M L2 I = 13/75 M L2 I = 19/147 M L2 I = 31/147 M L2 I = 19/192 M L2 I = 7/75 M L2 I = 7/81 M L2 I = 7/36 M L2 013 (part 2 of 3) 7.0 points If theta is in radians, then what is Newton's second law for rotational motion for this pendulum? Use the small angle a approximation. d2 theta /dt2 = -15/14 g/L theta d2 theta /dt2 = -72/43 g/L theta d2 theta /dt2 = -9/14 g/L theta d2 theta /dt2 = -63/38 g/L theta d2 theta /dt2 = -45/26 g/L theta d2 theta /dt2 = -12/7 g/L theta d2 theta /dt2 = -21/26 g/L theta d2 theta /dt2 =-105/62 g/L theta d2 theta /dt2 = -3/2 g/L theta d2 theta /dt2 = -24/19 g/L theta 014 (part 3 of 3) 7.0 points Based on the equation of the motion given in the previous question, the period of this pendulum in the small angle approximation is given by

Explanation / Answer

12)distance from the center=(4l/5)-l/2

=3l/10

so moment of inertia about the center=ml^2/12

applying parallel axis theorem,

I=ml^2/12 +(m*9l^2/100)

or I=13ml^2/75

13)-mg(4/5l-l/2)*theta=13/75*ml^2*alpha

or d2theta/dt^2=-(45/26)g/l *theta

14)d2t/dt^2=-w^2theta

so T=wpi*((26/25)*l/g)^1/2

12)4

13)5

14)2

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