An object is formed by attaching a uniform, thin rod with a mass of m_r = 7.38 k

Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 7.38 kg and length L = 4.84 m to a uniform sphere with mass m_s = 36.9 kg and radius R = 1.21 m. Note m_s = 5mr and L = 4R. What is the moment of inertia of the object about an axis at the left end of the rod? If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 421 N is exerted perpendicular to the rod at the center of the rod? 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 = 421 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? Compare the three moments of inertia calculated above: I_CM

Explanation / Answer

moment of inertia of sphere about its centre of mass, Icm = 2 mr^2/5

= (2 x 36.9 x 1.21^2)/ 5 = 21.61 kg m^2

about left end of rod , I1 = 26.61 + 36.9(4.84^2) = 886.01 kg m^2

moment of inertia of rod about left end, I2 = m L^2 / 3 = 57.63 kg m^3

Iet = I1 + I2 = 943.64 kg m^2

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2. torque = Ix alpha

(4.84/2) * (421) = 943.64 * alpha

alpha = 1.08 rad/s^2

3. of rod, I2 = 57.63 + (7.38 * (1.21/2)^2) = 60.33 kg m^3

of sphere = 26.61 + (36.9 * ( 1.21/2)^2 ) = 40.12 kg m^3

Inet = 60.33 + 40.12 = 100.45 kg m^3

4. torque = r x F = r Fsin@

@ - angle between r and F = 0
hence torque = 0

so alpha = 0

5. of rod, I2 = 57.63 + (7.38 * (1.21*2)^2) = 100.85 kg m^3

of sphere = 26.61 + (36.9 * ( 1.21)^2 ) = 80.64 kg m^3

Inet = I1 + I2= 181.5 kg m^3

6. Icm < Iright < Ileft

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