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

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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

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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

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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