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Set 8 Problem number 15
A net torque of 1346 meter Newtons is applied to a
massless disk constrained to rotate about an axis through its center and perpendicular to
its plane. Iron rods with mass density 13.35625 kilograms/meter, shaped into circles whose
radii are 10.49 meters, 20.98 meters and 31.47 meters, have been attached to the disk, concentric
with it.
- Find the angular acceleration of the system.
The first circle or hoop has circumference 2 `pi (
10.49 meters) = 65.8772 meters.
- Its mass is therefore 65.8772 meters ( 13.35625
kilograms/meter) = 879.8724 kg.
The second circle or hoop has circumference 2 `pi (
20.98 meters) = 131.7544 meters.
- Its mass is therefore 131.7544 meters ( 13.35625
kilograms/meter) = 1759.745 kg.
The third circle or hoop has circumference 2 `pi (
31.47 meters) = 197.6316 meters.
- Its mass is therefore 197.6316 meters ( 13.35625
kilograms/meter) = 2639.617 kg.
The three moments of inertia will therefore be
- hoop 1: ( 879.8724 kilograms)( 10.49 meters) ^ 2 = 290463.8
kg m ^ 2,
- hoop 2: ( 1759.745 kilograms)( 20.98 meters) ^ 2 = 774570
kg m ^ 2, and
- hoop 3: ( 2639.617 kilograms)( 31.47 meters) ^ 2 = 871391.3
kg m ^ 2.
These add up to the total moment of inertia
- total moment of inertia = I = 1936425 kg m ^ 2.
The 1346 meter Newton torque will result in an
angular acceleration of `alpha = `tau / I = 1346 meter Newtons / ( 1936425 kilogram meter ^
2) = 6.950953E-04 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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