Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 6.98 kg and length L = 5.04 m to a uniform sphere with mass m_s = 34.9 kg and radius R = 1.26 m. Note m_s = 5m_r 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

6)

Compare the three moments of inertia calculated above:  

I_CM < I_left < I_right

I_CM < I_right < I_left

I_right < I_CM < I_left

I_CM < I_left = I_right

I_right = I_CM < I_left

Answer 1

Answer:  m1=7,44 kg
L=5,04 m
m2=37,2 kg
R=1,26 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=15,749 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=23,623 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"=62,996 kg*m2
for the sphere we get J2"=J2+m2*sqr(d2) J2"=1500,091 kg*m2
And the total moment of inertia for the first case is
Jt1=J1"+J2"

FIRST ANSWER
Jt1=1563,087 kg*m2

(2)
F=488 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=1229,76 Nm
The acceleration can be found from
e1=M/Jt1

SECOND ANSWER
e1=0,787 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""=89,572 kg*m2
J2""=J2+m2*sqr(h2) J2""=82,682 kg*m2
and
Jt2=J1""+J2""

THIRD ANSWER
Jt2=172,254 kg*m2

(4)
F=488 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"""=204,737 kg*m2
J2"""=J2+m2*sqr(k2) J2"""=82,682 kg*m2
Jt3=J1"""+J2"""
So

THE FINAL ANSWER
Jt3=287,419 kg*m2

Answer 2

Given that,

mr =6.98 kg and L = 5.04m

ms = 34.9 kg and r = 1.26m

ms = 5mr and L = 4R

(1)the moment of inertia of the object about an axis at the left end of the rod

The moment of inertia for a rod rotating around its center is Jr=1/12m(r)²

Jr = 1/12(mr)(L)² = 1/12 (6.98)(5.04)² = 14.77 kg-m²  

The moment of inertia of a solid sphere rotating around its center will be Js = 2/5m(r)²

Js = 2/5(ms)(rs) = 2/5(34.9)(1.26)² = 22.16 kg-m²

As object 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 Jr"=Jr+mr (d1)² = Jr"= 59.09 kg-m²
for the sphere we get Js"=Js+ms(d2)² =  Js"= 1407.34 kg-m²
And the total moment of inertia for the first case is
J=Jr"+Js" = 59.09 + 1407 = 1466.431

J = 1466.431 kg-m²

(2) Angular acceleration to be calculated when F = 401 N is excerted

The torque given to a system in general is = Fd Sin (theta), where (theta) is the angle bet F and d = distance from rotating axis. Here (theta) = 90 degrees

Torque = 401 x 5.04/2 = 1010.52 Nm

Acceleration can be calculated from, a = Torque / J = 1010.52 / 1466.4 = 0.689 m/sec²

Hence a = 0.689 m/sec²

(3) moment of inertia of the object about an axis at the center of mass of the object

Again we have to use Steiner theorem
In this case h1=(L+R)/2
and h2=R/2
So
Jr""=J1+mr(h1)² = Jr""= 291.80 kg m²
Js""=J2+ms(h2)² = Js""= 36.011 kg m²
and
J' =J1""+J2"" = 291.80 + 36.011 = 327.8 kg m²

J' = 327.8 kg m²

(4) the angular acceleration if a force F = 401 N is exerted perpendicular to the rod at the center of the rod

Again, Torque = Fx(L+R)/2 = 401 x (5.04+1.26)/ 2 = 1263.15 N m

a = Torque / J' = 1263.15/ 327.8 = 3.8

a = 3.8 m/sec²