An object is formed by attaching a uniform, thin rod with a mass of m r = 7.13 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.13 kg and length L = 5.32 m to a uniform sphere with mass ms = 35.65 kg and radius R = 1.33 m. Note ms = 5mr and L = 4R.

1)

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

kg-m2

2)

If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 444 N is exerted perpendicular to the rod at the center of the rod?

rad/s2

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.)

kg-m2

4)

If the object is fixed at the center of mass, what is the angular acceleration if a force F = 444 N is exerted parallel to the rod at the end of rod?

rad/s2

5)

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

kg-m2

Explanation / Answer

for sphere ,

moment of inertia of sphere about its centreof mass axis Icm = (2 m r^2 / 5)

Icm = 2 x 35.65 x 1.33^2 / 5 = 25.22 kg m^2

moment of inertia of sphere about left end = Icm + md^2

I1 = 25.22 + ( 35.65 x (5.32+1.3)^2 ) = 1601.76 kg m^2

moment of inertia of rod about the left end I2 = mL^2 /3 = (7.13 x 5.32^2 / 3)

I2 = 67.26 kg m^2

Total moment of inertia = I1 + I2 = 1669.03 kg m^2

2) torque(r x F) = I x alpha

(5.32/2) x 444   = 1669.03 x alpha

alpha = 0.708 rad/s^2

3) for sphere ,

moment of inertia of sphere about its centreof mass axis Icm = (2 m r^2 / 5)

Icm = 2 x 35.65 x 1.33^2 / 5 = 25.22 kg m^2

moment of inertia of sphere about left end = Icm + md^2

I1 = 25.22 + ( 35.65 x (1.3/2)^2 ) = 40.28 kg m^2

moment of inertia of rod about that axis = mL^2 /12 + m((L+R)/2)^2 )

I2 = 7.13 x 5.32^2 /12   +   7.13 x ((5.32+1.33)/2)^2

I2 = 95.64

Total moment of inertia = I1 + I2 = 135.92 kg m^2

4) torque = r x F = rFsin0 = 0

5) for sphere ,

moment of inertia of sphere about its centreof mass axis Icm = (2 m r^2 / 5)

Icm = 2 x 35.65 x 1.33^2 / 5 = 25.22 kg m^2

moment of inertia of sphere about left end = Icm + md^2

I1 = 25.22 + ( 35.65 x (1.33)^2 ) = 90.05kg m^2

moment of inertia of rod = mL^2 /12 + m (L/2 + 2R)^2

I2 = 218.61 kg m^2

Total moment of inertia = I1 + I2 = 308.66 kg m^2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.