Question

An object is formed by attaching a uniform, thin rod with a mass of m_r = 7.25 kg and length L = 5.56 m to a uniform sphere with mass m_s = 36.25 kg and radius R = 1.39 m. Note m_s = 5m_r and L = 4R.

If the object is fixed at the left end of the rod, what is the angular acceleration if a force F = 481 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 = 481 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

M1 = 7.25 kg
length L = 5.56 m
mass M2 = 36.25 kg
Radius R = 1.39 m

sqr(x) means x*x

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

In this case J1=1/12*M1*sqr(L)

=18.676 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=28.015 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

d2=5.56 + 1.39= 6.95 m

For the rod it is d1=1/2*L=0.5*5.56=2.78m

From Steiner theorem

for the rod we get J1"=J1+m1* sqr(d1)

J1"=74.706 kg*m2

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

J2"=1778.98 kg*m2

And the total moment of inertia for the first case is

Jt1=J1"+J2"

Jt1=1853.68 kg*m2

F=481 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=1337.18 Nm

The acceleration can be found from

e1=M/Jt1

e1=0,721 rad/s2
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(3)

center of mass is.

Again we have to use Steiner theorem

In this case h1=(L+R)/2=(5.56+1.39)/2=3.475m

and h2=R/2=1.39/2=0.695

So

J1""=J1+m1*sqr(h1)
J1""=106.22 kg*m2

J2""=J2+m2*sqr(h2)

J2""=45.524 kg*m2

and

Jt2=J1""+J2""

Jt2=151.744 kg*m2
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(4)

F=481 N

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

M=0

and thus

e2=0 rad/s2
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