An object is formed by attaching a uniform, thin rod with a mass of m r = 6.93 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.93 kg and length L = 4.92 m to a uniform sphere with mass ms = 34.65 kg and radius R = 1.23 m. Note ms = 5mr and L = 4R.

Standard Exercise Tipler69P.045. -- Optional - An object is formed by attaching a uniform, thin rod with a mass of my = 6.93 kg and length L. 4.92 m to a uniform sphere with mass mg = .65 kg R = 1.23 m. 34and radius Note ms - Sm, and L = 4R. Standard Exercise Tipler6 9.P.049. Optional . I 1) What is the moment of inertia of the object about an axis at the left end of the rod? Standard Exercise Tipler69P.058. Optional . 1387.435 kg-mSubmit Help 2) If the object is fixed at the left end of the rod, what is the angular acceleration if a force F. 484 N is exerted perpendicular to the rod at the center of the rod? 0.858 radiSubmit Help What is the moment of inertia of the object about an axis at the center of mass of the object? (Note: the center of mass can be calculated to be located at a point halfway between the center of the sphere and the left edge of the sphere.) 113.6 ke-m? Submit Help 4) If the object is fixed at the center of mass, what is the angular acceleration if a force F. 484 N is exerted parallel to the rod at the end of rod? 0 radis2Submit Help O5) What is the moment of inertia of the object about an axis at the right edge of the sphere? 3075423 ke-m2 Submit

Explanation / Answer

1) moment of inertia of ball about axis passing through its centre = 2Ms R^2 / 5
     = 2 (34.65 x 1.23^2 ) /5 = 20.97

about the axis left end of rod = 20.97 + (34.65 x (4.92 + 1.23)^2 ) = 1331.52 kg m^2

moment of inertia of system = of rod + of sphere

I = ( 6.93 x 4.92^2 / 3 ) + ( 1331.52)

I = 1387.44 kg m^2

2) torque = r x F = I x alpha

(4.92 / 2) x 484 = (1387.44)alpha

alpha = 0.858 rad/s^2

3) distance of centre of mass of system from left end of rod = d

d = ( (6.93 x (4.92 /2 )) + ( 34.65 x (4.92 + 1.23)) ) / (6.92 + 34.65)

d = 5.54 m

using parallel axis theorem,

1387.44 = Icm + (6.93 +34.65)d^2

Icm = 112.97 kg m^2

4) torque = r x F = 0

as r and F are in same direction so cross product of r and F will be zero .

5)
now d = (1.23 +1.23 + 4.92) - 5.54 = 1.84 m

I = Icm + md^2

I = 112.97 + ((6.93 + 34.65) x 1.84^2) = 253.74 Kg. m^2

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