Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.1 kg and length L = 5.28 m to a uniform sphere with mass ms = 35.5 kg and radius R = 1.32 m. Note ms = 5mr and L = 4R.

I have the first couple questions right, but I can't figure out the next one.

The moment of inertia of the object about an axis at the left end of the rod is 1637.1 kg*m^2

If the object is fixed at the left end of the rod, the angular acceleration if a force F = 435 N is exerted perpendicular to the rod at the center of the rod is .7047

*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.)

If the object is fixed at the center of mass, the angular acceleration if a force F = 435 N is exerted parallel to the rod at the end of rod is zero.

What is the moment of inertia of the object about an axis at the right edge of the sphere?

Answer 1

Given that :

mass of the rod, m_r = 7.1 kg

length of the rod, L = 5.28 m

mass of the sphere, m_s = 35.5 kg

radius of the sphere, r = 1.32 m

(a) The moment of inertia of the object about an axis at the left end of the rod which is given as :

moment of inertia for rod at the center -

I_CM = (1/12) m_r L²                                                                     { eq.1 }

inserting the values in above eq.

I_CM = (1/12) (7.1 kg) (5.28 m)

I_CM = 16.5 kg.m²

moment of inertia for solid sphere at the center -

I_s = (2/5) m_s r²                                                                 { eq.2 }

inserting the values in eq.2,

I_s = (2/5) (35.5 kg) (1.32 m)²

I_s = 24.7 kg.m²

using a parallel axis theorem,

For a rod,   I_rod = I_CM + m_r (L/2)²                                                                      { eq.3 }

inserting the values in eq.3,

I_rod = (16.5 kg.m²) + (7.1 kg) [(5.28 m)/2]²

I_rod = (16.5 kg.m²) + (49.4 kg.m²)

I_rod = 65.9 kg.m²

For the sphere, we have

I_sphere = I_s + m_s (L + r)²                                                                      { eq.4 }

inserting the values in eq.4,

I_sphere = (24.7 kg.m²) + (35.5 kg) [(5.28 m) + (1.32 m)]²

I_sphere = (24.7 kg.m²) + (1546.3 kg.m²)

I_sphere = 1571 kg.m²

Now, the total moment of inertia is given by -

I_left = I_rod + I_sphere

I_left = (65.9 kg.m²) + (1571 kg.m²)

I_left = 1637 kg.m²

(2) If the object is fixed at the left end of the rod, then the angular acceleration will be given as :

using an equation,   = I

(L/2) F . Sin = I

where, F = 435 N   and = 90⁰

[(5.28 m)/2] (435 N) = (1637 kg.m²)

(1148.4 Nm) = (1637 kg.m²)

= 0.7 rad/s²

(3) The moment of inertia of the object about an axis at the center of mass of the object which is given as :

For rod, we have

I_rod' = I_CM + m_r [(L + R)/2]²    { eq.5 }

inserting the values in eq.5,

I_rod' = (16.5 kg.m²) + (7.1 kg) [(5.28 m) + (1.32 m)/2]²

I_rod' = (16.5 kg.m²) + (77.3 kg.m²)

I_rod' = 93.8 kg.m²

For sphere, we have

I_sphere' = I_s + m_s (r/2)²    { eq.6 }

inserting the values in eq.6,

I_sphere' = (24.7 kg.m²) + (35.5 kg) [(1.32 m)/2]²

I_sphere' = (24.7 kg.m²) + (15.4 kg.m²)

I_sphere' = 40.1 kg.m²

Now, the total moment of inertia is given by -

I_center = I_rod' + I_sphere'

I_center = (93.8 kg.m²) + (40.1 kg.m²)

I_center = 133.9 kg.m²

(4) If the object is fixed at the center of mass, then the angular acceleration will be given as :

torque may be defined as, = [(L + r)/2] F . Sin

where, = 180⁰

= 0 . it means that,   = 0