I need these 3 question please and thank you An object is formed by attaching a

Question

I need these 3 question please and thank you

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.86 kg and length L = 4.8 m to a uniform sphere with mass ms = 34.3 kg and radius R = 1.2 m. Note ms = 5mr and L = 4R.
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 = 465 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?

Explanation / Answer

3.

Moment of inertia of rod about its center of mass is

Ir,cm=(1/12)mrL2

From Parallel axis theorem for the rod gives

Ir=Ir,Cm +mr(L/2 +R/2)2

From Parallel axis theorem for the sphere gives

Is=Is,cm+ms(R/2)2=(2/5)msR2+(1/4)msR2 =(13/20)msR2

I=Ir+Is=(1/12)mrL2+mr(L/2 +R/2)2+(13/20)msR2

I=(1/12)*6.86*4.82 + 6.86*(4.8/2+1.2/2)2+(13/20)*34.3*1.22

I=107 Kg-m2

4)

T=FrSin(180) =0

so angular acceleration

alpha =T/I =0

5)

From parallel axis theorem

Is =(2/5)msR2+msR2 =(7/5)msR2

Ir =(1/12)mrL2+mr(2R+L/2)2

I=(1/12)mrL2+mr(2R+L/2)2+(7/5)msR2

I=(1/12)*6.86*4.82 + 6.86*(2*1.2+4.8/2)2+(7/5)*34.3*1.22

I=240.4 Kg-m2

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