Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 7.48 kg and length L = 5.04 m to a uniform sphere with mass m_s = 37.4 kg and radius R = 1.26 m. Note m_s = 5m_r and L = 4R.

1. What is the moment of inertia of the object about an axis at the left end of the rod?

2. If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 451 N is exerted perpendicular to the rod at the center of the rod?

3. 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.)

4. If the object is fixed at the center of mass, what is the angular acceleration if a force F = 451 N is exerted parallel to the rod at the end of rod?

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

6. Compare the three moments of inertia calculated above:  

a) I_CM < I_left < I_right

b) I_CM < I_right < I_left

c) I_right < I_CM < I_left

d) I_CM < I_left = I_right

e) I_right = I_CM < I_left

Answer 1

1)

we need to use parallel axis theorem

I will denote mass of sphere as m and mass of rod as M

I = I ( sphere) + I (rod)

I = m ( R + L)² + (2/5)m r² + 1/3ML²

I = 37.4 ( 1.26 + 5.04)² + (2/5) 37.4 * 1.26² + 1/3 * 7.48 * 5.04²

I = 1484.41 + 23.75 + 63.33

I = 1571.5 Kg.m²

_______________________________

2)

first, we find the torque applied by force

T = 451 * (5.04 / 2)

T = 1136.52 N.m

so,

angular acceleration = T / I

angular acceleration = 1136.52 / 1571.5

angular acceleration = 0.723 rad/sec²

_____________________

3)

I = (1/12)ML² + M ( L/2 + r/2)² + (2/5)mr² + m(r/2)²

I = (1/12) 7.48 * 5.04² + 7.48 * 3.15² + (2/5)37.4 * 1.26² + 37.4 * 0.63²

I = 15.83 + 74.22 + 23.75 + 14.844

I = 128.6 Kg.m²

______________________________

4)

torque = 0

so

angular acceleration = 0

___________________________

5)

for sphere, I = (7/5)mr² ( using parallel axis)

for rod, I = 1/12ML² + M ( 2r + L/2)²

so,

I = 1/12ML² + M ( 2r + L/2)² + (7/5)mr²

I = 15.83 + 190 + 83.126

I = 288.9 kg.m²

_______________________________

6)

I_CM < I_right < I_left