An object is formed by attaching a uniform, thin rod with a mass of mr = 6.72 kg

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.72 kg and length L = 5.52 m to a uniform sphere with mass ms = 33.6 kg and radius R = 1.38 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 = 421 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 = 421 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?

6)

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

m1=6.72 kg
L=5.524 m
m2=33.6 kg
R=1.38 m

sqr(x) means x*x
(1)
The moment of inertia for a rod rotating around its center is J1=1/12*m*sqr(r)
In this case J1=1/12*m1*sqr(L) J1=17.06 kg*m2
The moment of inertia for a solid sphere rotating around its center is J2=2/5*m*sqr(r)
In this case J2=2/5*m2*sqr(R) J2=25.595 kg*m2

As the thigy rotates around the free end of the rod then for the sphere the axis around what it rotates is at a distance of d2=L+R
For the rod it is d1=1/2*L

From Steiner theorem
for the rod we get J1"=J1+m1* sqr(d1) J1"=68.25 kg*m2
for the sphere we get J2"=J2+m2*sqr(d2) J2"=1625.29 kg*m2
And the total moment of inertia for the first case is
Jt1=J1"+J2"

FIRST ANSWER
Jt1=1625.29 kg*m2

(2)
F=421 N
The torque given to a system in general is
M=F*d*sin(a) where a is the angle between F and d
and where d is the distance from the rotating axis. In this case a=90" and so
M=F*L/2 M=1161.96 Nm
The acceleration can be found from
e1=M/Jt1

SECOND ANSWER
e1=0.686 rad/s2

(3)
I assume the text to be right in the case where the center of mass is.
Again we have to use Steiner theorem
In this case h1=(L+R)/2 =3.45
and h2=R/2 =0.69
So
J1""=J1+m1*sqr(h1) J1""=97.04 kg*m2
J2""=J2+m2*sqr(h2) J2""=41.58 kg*m2
and
Jt2=J1""+J2""

THIRD ANSWER
Jt2=138.62 kg*m2

(4)
F=421 N
M=F*(L+R/2)*sin(a) In this case a=0" and so
M=0
and thus

FOURTH ANSWER
e2=0 rad/s2

(5)
In this case again we have to use Steiner theorem
k1=2*R+L/2 =19.32
K2=R =1.38

so
J1"""=J1+m1*sqr(k1) J1"""=2525.38 kg*m2
J2"""=J2+m2*sqr(k2) J2"""=89.577 kg*m2
Jt3=J1"""+J2"""
So

THE FINAL ANSWER
Jt3=2614.96 kg*m2

6. Icm < Ileft <  Iright ( Comparing the values)

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