An object is formed by attaching a uniform, thin rod with a mass of m r = 6.85 k

Question

An object is formed by attaching a uniform, thin rod with a mass of mr = 6.85 kg and length L = 5.76 m to a uniform sphere with mass ms = 34.25 kg and radius R = 1.44 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 = 460 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 = 460 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

mass of mr = 6.85 kg

length L = 5.76 m to

uniform sphere with mass ms = 34.25 kg

radius R = 1.44 m.

(1)The moment of inertia for a rod rotating around its center is J1=1/12*m*r2

In this case J1=1/12*m1*L2

J1=18.94 kg*m2

The moment of inertia for a solid sphere rotating around its center is

J2=2/5*m*r2

In this case J2=2/5*m2*R2

J2=28.41 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* d12

J1'=75.76 kg*m2

for the sphere we get J2'=J2+m2*d22

J2'=1803.9 kg*m2

And the total moment of inertia for the first case is

Jt1=J1'+J2'

Jt1=1979.7 kg*m2

b)F=460 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=1324.8 Nm

The acceleration can be found from

e1=M/Jt1

e1=0.669 rad/s2

(3)

We can use Steiner theorem

In this case h1=(L+R)/2

and h2=R/2

So

J1''=J1+m1*h12

J1''=107.616 kg*m2

J2''=J2+m2*h22

J2''=46.16 kg*m2

and

Jt2=J1"+J2"

Jt2=153.78 kg*m2

(4)

F=460 N

M=F*(L+R/2)*sin(a) In this case a=0" and so

M=0

and thus

e2=0 rad/s2

(5)

Let us use the Steiner theorem

k1=2*R+L/2

K2=R

J1"'=J1+m1*k12

J1'''=246.2 kg*m2

J2'''=J2+m2*k2

J2'''=99.43 kg*m2

Jt3=J1'''+J2'''

So

Jt3=345.6 kg*m2

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