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Set 8 Problem number 15
A massless disk is constrained to rotate about an
axis through its center and perpendicular to its plane. Iron rods are shaped into circles
whose radii are 9.6 meters, 19.2 meters and 28.8 meters. These circles are secured to the disk,
concentric with it. The rods have mass density 9.167304 kilograms/meter.
- Find the angular acceleration that will result from
the application of a torque of 1879 meter Newtons.
The first circle or hoop has circumference 2 `pi (
9.6 meters) = 60.28801 meters.
- Its mass is therefore 60.28801 meters ( 9.167304
kilograms/meter) = 552.6785 kg.
The second circle or hoop has circumference 2 `pi (
19.2 meters) = 120.576 meters.
- Its mass is therefore 120.576 meters ( 9.167304
kilograms/meter) = 1105.357 kg.
The third circle or hoop has circumference 2 `pi (
28.8 meters) = 180.864 meters.
- Its mass is therefore 180.864 meters ( 9.167304
kilograms/meter) = 1658.035 kg.
The three moments of inertia will therefore be
- hoop 1: ( 552.6785 kilograms)( 9.6 meters) ^ 2 = 152804.5
kg m ^ 2,
- hoop 2: ( 1105.357 kilograms)( 19.2 meters) ^ 2 = 407478.8
kg m ^ 2, and
- hoop 3: ( 1658.035 kilograms)( 28.8 meters) ^ 2 = 458413.7
kg m ^ 2.
These add up to the total moment of inertia
- total moment of inertia = I = 1018697 kg m ^ 2.
The 1879 meter Newton torque will result in an
angular acceleration of `alpha = `tau / I = 1879 meter Newtons / ( 1018697 kilogram meter ^
2) = 1.844513E-03 radians/second.
A hoop of radius r and mass density `lambda,
measured in kg / meter, will have circumference 2 `pi r and therefore total mass hoop mass
= 2 `pi r * `lambda.
- If the hoop rotates about an axis through its center
and perpendicular to its plane, then its entire mass lies at the same distance from the
axis of rotation; the entire mass moves as a unit and therefore has moment of inertia
I = m r^2 = 2 `pi r * `lambda * r^2 = 2 `pi r^3 *
`lambda.
A series of hoops with radii r1, r2, ..., rn, each
with the same density `lambda, will have circumferences
- hoop circumferences: 2 `pi r1, 2 `pi r2,
..., 2 `pi rn,
masses
- hoop masses: 2 `pi r1 * `lambda, 2 `pi
r2 * `lambda, ..., 2 `pi rn * `lambda,
and will therefore have moments of inertia
- hoop moments of inertia: 2 `pi r1^3 *
`lambda, 2 `pi r2^3 * `lambda, ..., 2 `pi rn^3 * `lambda.
The total moment of inertia will be
- I = 2 `pi r1^3 * `lambda `2 `pi r2^3 * `lambda 0...
`2 `pi rn^3 * `lambda = 2 `pi `lambda * ( r1^3 + r2^3 + ... + rn^3).
The acceleration resulting from applying a torque
`tau will therefore be
- angular acceleration = `tau / I = `tau / [ 2 `pi
`lambda * ( r1^3 + r2^3 +... + rn^3) ].
If mass m is distributed over a circle or a hoop of
radius r centered at the axis of rotation then the entire mass m lies at distance r from
the axis and the moment of inertia of that hoop is m r^2.
- If we have n circular hoops constrained to rotate
together, with masses m1, m2, ..., mn and radii r1, r2, ..., rn their total moment of
inertia is I = `Sigma ( m r^2) = m1 r1^2 + m2 r2^2 + ... + mn rn^2.
- If this system is subjected to net torque `tau it
will have angular acceleration
- `alpha = `tau / I = `tau / (m1 r1^2 + m2 r2^2 + ...
+ mn rn^2).
The figure below depicts three concentric hoops.
- Each hoop has its own mass and radius and therefore
its own moment of inertia.
- The moment of inertia of the system is the sum of
the individual moments of inertia.
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