An object is formed by attaching a uniform, thin rod with a mass of m r = 6.78 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.78 kg and length L = 5.64 m to a uniform sphere with mass ms = 33.9 kg and radius R = 1.41 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?

kg-m2  

2)

If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 441 N is exerted perpendicular to the rod at the center of the rod?

rad/s2  

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.)

kg-m2  

4)

If the object is fixed at the center of mass, what is the angular acceleration if a force F = 441 N is exerted parallel to the rod at the end of rod?

rad/s2  

5)

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

kg-m2

6)

Compare the three moments of inertia calculated above:

ICM < Ileft < Iright

ICM < Iright < Ileft

Iright < ICM < Ileft

ICM < Ileft = Iright

Iright = ICM < Ileft

What is the moment of inertia of the object about an axis at the right edge of the sphere? kg-m2 6) Compare the three moments of inertia calculated above: ICM

Explanation / Answer

m1=6.78 kg
L=5.64 m
m2=33.9 kg
R=1.41 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=17.97249 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=26.958 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"=71.889 kg*m2

for the sphere we get J2"=J2+m2*sqr(d2)

J2"=1711.87 kg*m2

And the total moment of inertia for the first case is
Jt1=J1"+J2"

FIRST ANSWER
Jt1=1783.76 kg*m2   

(2)
F=441 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=1243.62 Nm

The acceleration can be found from

e1=M/Jt1

SECOND ANSWER
e1=0.6971 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
and h2=R/2
So
J1""=J1+m1*sqr(h1)

J1""=102.21 kg*m2

J2""=J2+m2*sqr(h2)

J2""=43.80 kg*m2
and
Jt2=J1""+J2""

THIRD ANSWER
Jt2=146.017 kg*m2

(4)
F=441 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
K2=R
so
J1"""=J1+m1*sqr(k1)

J1"""=17.97249 + 215.669 = 233.64 kg.m2

J2"""=J2+m2*sqr(k2)

J2"""=26.958 + 67.39659 = 94.35 kg.m2

Jt3=J1"""+J2"""
So
THE FINAL ANSWER
Jt3 = 327.99459 kg.m2

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