ENGINEERING PHYSICS SECTION B1 : rigid bodies, rotational dynamics

Moment of inertia is the name given to rotational inertia, the rotational analogue for linear motion. Linear inertia is mass, or resistance to motion in a straight line. Rotational inertia of a body is a measure of its resistance to motion about some circular axis. It appears in the relationships for the dynamics of rotational motion.

The moment of inertia must be specified with respect to a chosen axis of rotation (see diagram).

For a point mass, the moment of inertia, I, is just the mass times the square of perpendicular distance to the rotation axis, so

I (kgm2)=mr 2

That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection of point masses. So, we can write:

I=Σmr2

where Σ=the∑ of allthe point masses at

their respectivevalues of r

If required, you’ll be given the relevant MoI for a particular body – here’s a few examples. They’re not too difficult to derive if you can integrate a bit, but we don’t need to here. Notice that M (mass) and R for radius, and W or L for length appear in them all.

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ENGINEERING PHYSICS SECTION B1 : rigid bodies, rotational dynamics

Example:

I have four identical masses m arrayed about an origin. Mass 1 is located at (1,1) mass 2 is located at (-1,1) mass 3 is located at (-1,-1) mass 4 is located at (1,-1)

Find the moment of inertia about the origin.

Clearly, the centre of mass is located at (0,0). Thus, each mass m is √2units (r) from the origin. I=Σmr2 = mx (√2 squared )for each, so I=4 xm x2 = 8m

Rotational vs Linear:

The rotational equivalent of Newton' s second lawis therefore

ΣF=rate of changeof linear momentum=m dvdt

=ma

is∑Γ=rate of changeof angular momentum=I dω

dt=I a

This from Khan Academy:

https://www.khanacademy.org/science/physics/torque-angular-momentum/torque-tutorial/v/rotational-version-of-newtons-second-law.

Examples:

John MacArthur 2

ENGINEERING PHYSICS SECTION B1 : rigid bodies, rotational dynamics

For a fixed input of rotational kinetic energy: the skater approximates to a cylinder. By increasing the radius from the axis of rotation, the moment of inertia increases thus slowing down the speed of rotation. Alternatively, if the skater wants to increase the speed of rotation, or angular velocity, then they must decrease the radius by bringing the segments of the body closer to the axis of rotation thus decreasing the radius and moment of inertia.

Worked example:

A circular saw blade is in the form of a flat disc, mass 2kg and radius 40cm.When supplied with a 2.4 kW electrical supply, assuming no loss of energy, what will its final angular velocity be? Treat the blade as a solid cylinder.

Check the working with me.

Electrical energy input = angular kinetic energy output

2400=12I ω2

where (see above) for a cylinder:I=12M R2

from which ω=√ 4 x24002 x0.42

=173.2rad s−1≅ 28rev s−1=1654 rpm

(since2π rad=3600=1 rev)

John MacArthur 3

ENGINEERING PHYSICS SECTION B1 : rigid bodies, rotational dynamics

This would make a very high-pitched squeal which we wouldn’t hear in practice, thus the energy transfer is not efficient.

If the actual angular velocity achieved was 75 rad s-1, what is the efficiency of the motor?

If it takes 8.0s to achieve this angular velocity from rest, find its angular acceleration

efficiency= poweroutpower∈¿¿

soω2=752=4 x Pout2x 0.42

so Pout=450W

eff= 4502400

=18.8%

sinceα= ΔωΔt

=ω f−ωit

=75−08

=9.38 rad s−2(don' t convert ¿ rpmshere)

John MacArthur 4