An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.83 k

Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.83 kg and length L = 5.72 m to a uniform sphere with mass m_s = 34.15 kg and radius R = 1.43 m. What is the moment of inertia of the object about an axis at the left end of the rod? 650 kg-m^2 If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 454 N is exerted perpendicular to the rod at the center of the rod? rad/s^2 What is the moment of inertia of the object about an axis at the center of mass of the object? If the object is fixed at the center of mass, what is the angular acceleration if a force F = 454 N is exerted parallel to the rod at the end of rod? What is the moment of inertia of the object about an axis at the right edge of the sphere? kg-m^2

Explanation / Answer

1) I_(sphere)cm = 2 ms R^2 / 5

= 2 x (5 m ) R^2 / 5

= 2 m R^2

moment of inertia of sphere about the axis,

I1 = Icm + m d^2

I1 = (2 m R^2) + (ms)(R + L)^2

I1 = 2 m R^2 + (5 m )(5 R)^2

I1 = 127 m R^2

moment of inertia of rod, I2 = m L^2 / 3 = m (4 R)^2 / 3 = 5.33 m R^2

I = I1 + I2 =132.22 m R^2

I = (132.22) (6.83) (1.43^2) = 1846.7 kg m^2

(2) torque = r x F = (2 R) F = 2 x 1.43 x 454 = 1298.44 N m

angular acceleration = torque / I

= 1298.44 / 1846.7

= 0.703 rad/s^2

(3) for sphere,

I1 = ( 2 m R^2) + (5 m)( 0.5 R)^2 = 3.25 m R^2

for rod:

Icm = m L^2 / 12 = 4 m R^2 / 3

I2 = (4 m R^2 / 3) + (m (2.5R)^2)

I2 = 7.58 m R^2

I = I1 + I2 = 10.83 m R^2

I = (10.83) (6.83) (1.43^2) = 151.3 kg m^2

(4) zero.

(5) for sphere,

I1 = ( 2 m R^2) + (5 m)( R)^2 = 7 m R^2

for rod:

Icm = m L^2 / 12 = 4 m R^2 / 3

I2 = (4 m R^2 / 3) + (m (4R)^2)

I2 = 17.33 m R^2

I = I1 + I2 = 24.33 m R^2

I = (24.33) (6.83) (1.43^2) = 340 kg m^2

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