1- What is the moment of inertia of the object about an axis at the left end of

Question

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 = 455 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 = 455 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?

Moment of Inertia 12 34 5 6 7 1 2 345 6 7 An object is formed by attaching a uniform, thin rod with a mass of mr- 6.83 kg and length L 5.72 m to a uniform sphere with mass m 34.15 kg and radius R-1.43 m. Note ms 5mr and L- 4R.

Explanation / Answer

remember that we can only answer a certain amount of subquestions, please repost for the missing ones

the inertia of a rod around its center is mL^2/12 and the moment of a sphere is 2mr^2/5

using the parallel axis theorem we have that:

for left edge:

mL^2/12+mL^2+ 2MR^2/5+M(L+R)^2

taking the relations

I=m(4R)^2/12+m(4R)^2+2mR^2+5m(5R)^2

I=6.83*(5.72)^2/12+6.83*(5.72)^2+2*6.83*1.43^2+5*6.83*(5*1.43)^2

I=2015.85 kgm^2

for right edge:

I=m(4R)^2/12+m(2R+L/2)^2++2mR^2+5mR^2

I=6.83*(5.72)^2/12+6.83*(2.86+5.72/2)^2++2*6.83*1.43^2+5*6.83*1.43^2

I=339.85 kgm^2

In both case we can use newton's second law of rotation

Torque=I*alpha

alpha=torque/I

For the left end

alpha=-455*5.72/(2*2015.85)

alpha=- 0.645 rad/s^2

for the question 4 the force is parallel to the rod so the torque is zero and ergo the angular acceleration is also zero

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