An object is formed by attaching a uniform, thin rod with a mass of m r = 7.24 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.24 kg and length L = 5.56 m to a uniform sphere with mass ms = 36.2 kg and radius R = 1.39 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?

2) If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 479 N is exerted perpendicular to the rod at the center of the rod?

3)

4) If the object is fixed at the center of mass, what is the angular acceleration if a force F = 479 N is exerted parallel to the rod at the end of rod?

Explanation / Answer

a) I = ml^2/3 + 4Mr^2/5 + M(l+r)^2 = 1879.108937 kgm^2
b) 479*5.56/2 = I*alpha
alpha = 0.708644 rad/s/s
c) COm, x = (7.24*5.56/2 + 36.2*(5.56+1.39))/(7.24+36.2) = 6.255 m
MOment of inertia = ml^2/12 + m(x - l/2)^2 + 4*M*r^2/5 + M(r/2)^2 = 179.51785 kg m^2
d) Torque = F x R = 0

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