Question

Moment of Inertia

To find the moment of inertia of different objects and to observe the changes in angular acceleration relative to changing moments of inertia.

To also learn how to use calipers in making precise measurements

The momentum of inertia of an object is calculated as I=∑mr^2

If the object in question rotates around a central point, then it can be considered a "point mass", and its moment of inertia is simply, I=mr^2 where r is from the central point to the center of mass.

If you have a mass (m = 1.75 kg, any shape or size) that is revolving around in a circle (radius r = 15.0 cm), what is its moment of inertia? I = ? kg*m2

If you have more than one object revolving around a point, then the total moment of inertia is the sum of the two: I_Total=I_1+I_2+I_3+…

If the object itself is rotating around the central point (central point inside the mass), then the moment of inertia is the sum of mr^2 of all possible points from the central point. Normally, you need calculus to do this sum, but the results are just given to you in the textbook (Figure 8-20 in the textbook).

Here are the moments of inertia for three object shapes: (For the rod, this is a rod rotating around its center, like a helicopter's rotors.)

I_rod=1/12 ML^2 I_disk=1/2 MR^2 I_ring=1/2 M(R_inner^2+R_outer^2)

Calculate the moment of inertia of a rod (m = 1.20 kg, L = 45 cm) that is revolving around its center. I = ? kg*m2

Finding the total moment of inertia of an object, just find the I for each object, then just take the sum to find the total.

You have a rod (m = 1.20 kg, L = 45 cm) that is revolving around its center, with two individual masses (m = 1.75 kg) attached 15.0 cm from the rods center. Calculte the total moment of inertia. I = ? kg*m2

Calculate the moment of inertia of a solid wheel (disk) with a mass of m = 7.00 kg and a diameter of 90 cm. I = ? kg*m2

Calculate the moment of inertia of a small ring with a mass of m = 50.0 g, an inner radius of 0.50 cm and an outer radius of 0.52 cm. I = ? kg*m2

Answer #1

**moment of inertia of rod I_rod = (1/12)*mrod*L^2 =
(1/12)*1.2*0.45^2 - 0.02025 kg m^2**

**---------------------------**

**inertia of object = (1/12)*mrod*r^2 + 2*m*particle*l^2 =
0.02025 + 2*1.75*0.15^2 = 0.099 kg m^2**

**----------------------------------**

**Iwheel = (1/2)*mwheel*R^2**

**R r adius = 90/2 = 45 cm = 0.45 m**

**Iwheel = (1/2)*7*0.45^2 = 0.70875 kg m^2**

**========================**

**Iring = M(R_inner^2+R_outer^2)**

**R_inner = 0.5/100 = 0.005 m**

**R_outer = 0.52 cm = 0.0052 m**

**Iring = 50*10^-3*(0.005^2+0.0052^2)**

**Iring = 2.602*10^-6 kg m^2**

An object is formed by attaching a uniform, thin rod with a mass
of mr = 7.22 kg and length L = 5.52 m to a uniform
sphere with mass ms = 36.1 kg and radius R = 1.38 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?
2)If the object is fixed at the left end of the rod, what...

A rod of length l=1.1m and mass M= 5.5kg joins two particles
with masses m1 =4.8kg and m2 = 2.8kg, at its ends. The combination
rotates in the xy-plane about a pivot through the center of the rod
with the linear speed of the masses of v= 3.5 m/s. (Moment of
inertia of a uniform rod rotating about its center of mass I= 1 12
M l2 ) angularmomentum a) Calculate the total moment of inertia of
the system I...

A rod of length l=2.2m and mass M= 9.7kg joins two particles
with masses m1 =12.9kg and m2 = 5.0kg, at its
ends. The combination rotates in the xy-plane about a pivot through
the center of the rod with the linear speed of the masses of v=
12.9 m/s. (Moment of inertia of a uniform rod rotating about its
center of mass
I=
1
12
M l2
)
a) Calculate the total moment of inertia of the system
I =...

An object with moment of inertia ?1 = 9.7 ? 10−4 ?? ∙ ?2 rotates
at a speed of 3.0 ???/?. A 20 ? mass with moment of inertia ?2 =
1.32 ? 10−6 ?? ∙ ?2 is dropped onto the rotating object at a
distance of 5.0 ?? from the center of mass. What is the angular
velocity of the combined object and mass after the drop?

1) A torque of 1.20 N m is applied to a thin rod of mass 2.50 kg
and length 50.0 cm pivoted about its center and at rest. How fast
is the rod spinning after 4.25 s?
a. 32.6 rad/s
b. 8.16 rad/s
c. 97.9 rad/s
d. 24.5 rad/s
2) A torque of 1.20 N m is applied to a thin rod of mass 2.50 kg
and length 50.0 cm pivoted about one end and at rest. How fast is...

An object with moment of inertia ?1 = 9.7 ? 10−4 ?? ∙ ?2 rotates
at a speed of 3.0 ???/?. A 20 ? mass with moment of inertia ?2 =
1.32 ? 10−6 ?? ∙ ?2 is dropped onto the rotating object at a
distance of 5.0 ?? from the center of mass. What is the angular
velocity of the combined object and mass after the drop?
Please don't forget to state the combined mass after the
drop!!!

Three objects: disk, cylinder, and sphere are each rotating at 5
rads/s about an axis through their center. If the mass and radius
of each object is 5 kg and 2 m respectively.
(a) What is is the moment of inertia of each object?
(b) What is is the angular momentum of each object?

Find the moment of inertia about each of the following axes for
a rod that has a diameter of d, a length of l, and a mass of m. A)
About an axis perpendicular to the rod and passing through its
center. B) About an axis perpendicular to the rod and passing
through one end. C) About a longitudinal axis passing through the
center of the rod.

THE BUBBLE MOMENT
A physical pendulum is composed a rod of length 26 cm and radius
3.2 cm suspended downward from one of its ends. At the bottom, free
end, there is a large, circular bubble of the same radius as the
rod so that it is wholly within the rod, but as far towards the
bottom as possible.
(a) If the rod has a density of 9 g/cm3, what is its
mass?
(b) Find its center of mass if...

A playground merry-go-round has a radius of 3m and a moment of
inertia of 1500 kg m^2. 4 kids sit at the outer edge with each kid
having a mass of 20 kg, and the merry-go-round spins at 1.75 rad/s.
If 2 of the 4 kids move to a position that is 0.3m from the center
of the merry-go-round, what is the new rotational speed of the
merry-go-round? No energy is lost to friction and the children are
point masses.

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