An object is formed by attaching a uniform, thin rod with a mass of m r = 6.64 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.64 kg and length L = 5.36 m to a uniform sphere with mass ms = 33.2 kg and radius R = 1.34 m. Note ms = 5mr and L = 4R.

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

b) If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 498 N is exerted perpendicular to the rod at the center of the rod?

c) 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.)

d) If the object is fixed at the center of mass, what is the angular acceleration if a force F = 498 N is exerted parallel to the rod at the end of rod?

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

f) Compare the three moments of inertia calculated above:

ICM < Ileft < Iright

ICM < Iright < Ileft

Iright < ICM < Ileft

ICM < Ileft = Iright

Iright = ICM < Ileft

Explanation / Answer

given data

mr = 6.64 kg, L = 5.36 m, ms = 33.2 kg, R = 1.34 m

1)

I_left = mr*L^2/3 + (2/5)*ms*R^2 + ms*(L + R)^2

= 6.64*5.36^2/3 + (2/5)*33.2*1.34^2 + 33.2*(5.36 + 1.34)^2

= 1578 kg.m^2

2) Torque due to the force, T = r*F

= (5.36/2)*498

= 1335 N.m

angular acceleration, alfa = T/I

= 1335/1578

= 0.846 rad/s^2

3) let x is the distance between center of mass to center of the sphere.

ms*x = mr*(L/2 + R-x)

33.2*x = 6.64*(5.36/2 + 1.34 - x)

==> x = 0.67 m

Icm = (2/5)*ms*R^2 + ms*x^2 + mr*L^2/12 + ms*(L/2 + R - x)^2

= (2/5)*33.2*1.34^2 + 33.2*0.67^2 + 6.64*5.36^2/12 + 6.64*(5.36/2 + 1.34 - 0.67)^2

= 129.2 kg.m^2

4) T = r*F

= (L/2 + R - x)*F

= (5.36/2 + 1.34 - 0.67)*498

= 1668 N

angular acceleration, alfa = T/I

= 1668/129.2

= 12.9 rad/s^2

5) I_right = mr*L^2/12 + mr*(L/2 + 2*R)^2 + (2/5)*ms*R^2 + ms*(2*R)^2

= 6.64*5.36^2/12 + 6.64*(5.36/2 + 2*1.34)^2 + (2/5)*33.2*1.34^2 + 33.2*(2*1.34)^2

= 469 kg.m^2

6) Icm < I_right < I_left

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