An object is formed by attaching a uniform, thin rod with a mass of m r = 7.32 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.32 kg and length L = 5.72 m to a uniform sphere with mass ms = 36.6 kg and radius R = 1.43 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 = 401 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 = 401 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

Explanation / Answer

m1=7.32 kg
L=5.72 m
m2=36.6 kg
R=1.43 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=1/12*7.32*5.722 =19.958 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=2/5*36.6*1.432 =29.937 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) = 19.958 + 7.32*(5.72/2)2 = 79.83 kg*m2

for the sphere we get J2"=J2+m2*sqr(d2) = 29.937 + 36.6*(5.72+1.43)2 = 1901.02 kg*m2
And the total moment of inertia for the first case is
Jt1=J1"+J2"

FIRST ANSWER
Jt1=1980.85 kg*m2

(2)
F=401 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=401*5.72/2 = 1146.86 Nm
The acceleration can be found from
e1=M/Jt1

SECOND ANSWER
e1=0.5789 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""= 19.958 + 7.32*((5.72+1.43)/2)2 = 113.51 kg*m2
J2""=J2+m2*sqr(h2)

J2""= 29.937 + 36.6*(1.43/2)2 =48.64 kg*m2
and
Jt2=J1""+J2""

THIRD ANSWER
Jt2=162.16 kg*m2

(4)
F=401 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)
J2"""=J2+m2*sqr(k2)
Jt3=J1"""+J2"""

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