n object is formed by attaching a uniform, thin rod with a mass of mr = 7.11 kg

Question

n object is formed by attaching a uniform, thin rod with a mass of mr = 7.11 kg and length L = 5.28 m to a uniform sphere with mass ms = 35.55 kg and radius R = 1.32 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 = 438.0 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 = 438.0 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

Explanation / Answer

m1=7.1 kg
L=5,28 m
m2=35.5 kg
R=1,32 m

sqr(x) means x*x
(1)
The moment of inertiafor a rod rotating around its center is J1=1/12*m*sqr(r)
In this case J1=1/12*m1*sqr(L) J1=16.49 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=24.74 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"=65.97 kg*m2

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

J2"=1571.13 kg*m2

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

FIRST ANSWER
Jt1=1637.1 kg*m2

(2)
F=437 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=1153.68 Nm
The acceleration can be found from
e1=M/Jt1

SECOND ANSWER
e1=0.7047 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 = 3.3
and h2=R/2 = 0.66
So
J1""=J1+m1*sqr(h1).... J1""=......
J2""=J2+m2*sqr(h2).... J2""........

and
Jt2=J1""+J2""



(4)
F=437 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"""=..................
J2"""=J2+m2*sqr(k2) J2"""=................
Jt3=J1"""+J2"""

Related Questions

Navigate

• Browse (All)
• Browse N
• Subjects

---

---

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