An object is formed by attaching a uniform, thin rod with a mass of m r = 7.36 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 7.36 kg and length L = 4.8 m to a uniform sphere with mass ms = 36.8 kg and radius R = 1.2 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 = 414 N is exerted perpendicular to the rod at the center of the rod? rad/s2

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 = 414 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:

a)ICM < Ileft < Iright

b)ICM < Iright < Ileft

c)Iright < ICM < Ileft

d)ICM < Ileft = Iright

e)Iright = ICM < Ileft

Explanation / Answer

m1=7.36 kg
L=4.8 m
m2=36.8 kg
R=1.2 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=14.1312 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=21.1968 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"=56.5248 kg*m2
for the sphere we get J2"=J2+m2*sqr(d2) ;J2"=1345.9968 kg*m2
And the total moment of inertia for the first case is
Jt1=J1"+J2"

FIRST ANSWER
Jt1=1402.5216 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=1171.2 Nm
The acceleration can be found from
e1=M/Jt1

SECOND ANSWER
e1=0.835 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""=80.3712 kg*m2
J2""=J2+m2*sqr(h2) J2""=34.4448 kg*m2
and
Jt2=J1""+J2""

THIRD ANSWER
Jt2=114.816 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"""=183.7056 kg*m2
J2"""=J2+m2*sqr(k2) J2"""=74.1888 kg*m2
Jt3=J1"""+J2"""
So

THE FINAL ANSWER
Jt3=257.8944 kg*m2

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.