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Set 8 Problem number 14
What angular acceleration results when a net torque
of 1862 meter Newtons is applied to the system described below?
The system consists of a massless disk is
constrained to rotate about an axis through its center and perpendicular to its plane.
- Circles concentric with the disk are drawn on the
disk, the first having radius 3.1 meters, the second twice and the third three times that
radius.
- Around each circle, masses of 7 kilograms are
evenly distributed, with 1.033333 / `pi meters of arc between masses.
The first circle has a total circumference of 2 `pi
( 3.1 meters).
- With masses spaced at `pi / 1.033333 meters, the number of
masses on the first circle will be
- number of masses on first circle = circumference /
arc distance between masses = 2 `pi ( 3.1 meters)/( `pi / 1.033333 meters) = 2( 3.1 )( 1.033333 ) =
6.406666 .
- Noting that the second and third circles have double
and triple the circumference of the first, with masses spaced identically to the first,
there will be 2( 6.406666 ) and 3( 6.406666 ) masses on the second third circles.
At 7 kilograms per mass, the circles will thus
have masses
- Circle 1 total mass: m1 = 44.84666 kilograms
- Circle 2 total mass: m2 = 89.69333 kilograms
- Circle 3 total mass: m3 = 134.54 kilograms
The moments of inertia of the circles are thus
- Circle 1 moment of inertia: m1 * r1^2 = ( 44.84666 kg)(
3.1 meters) ^ 2 = 139.0247 kg m ^ 2,
- Circle 2 moment of inertia: m2 * r2^2 = ( 89.69333 kg)(
6.2 meters) ^ 2 = 278.0493 kg m ^ 2, and
- Circle 3 moment of inertia: m3 * r3^2 = ( 134.54 kg)(
9.299999 meters) ^ 2 = 417.074 kg m ^ 2.
The total moment of inertia is therefore the sum
834.1479 kg m^2 of these mr ^ 2 contributions.
A torque of 1862 meter Newtons will thus result in an
acceleration of
- angular acceleration = `alpha = `tau / I = 1862 meter
Newtons / 834.1479 kg m ^ 2 = 2.232218 rad/second ^ 2.
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 masses distributed
uniformly over three concentric hoops.
- The moment of inertia of a hoop of radius r
containing n masses, each of mass m, will be I = (hoop mass) * r^2 = n * m r^2.
- The total moment of inertia will be the total of the
moments for the individual hoops.
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