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Set 8 Problem number 14
What angular acceleration results when a net torque
of 1235 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 6.6 meters, the second twice and the third three times that
radius.
- Around each circle, masses of 7 kilograms are
evenly distributed, with 1.65 / `pi meters of arc between masses.
The first circle has a total circumference of 2 `pi
( 6.6 meters).
- With masses spaced at `pi / 1.65 meters, the number of
masses on the first circle will be
- number of masses on first circle = circumference /
arc distance between masses = 2 `pi ( 6.6 meters)/( `pi / 1.65 meters) = 2( 6.6 )( 1.65 ) =
21.78 .
- 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( 21.78 ) and 3( 21.78 ) masses on the second third circles.
At 7 kilograms per mass, the circles will thus
have masses
- Circle 1 total mass: m1 = 152.46 kilograms
- Circle 2 total mass: m2 = 304.92 kilograms
- Circle 3 total mass: m3 = 457.38 kilograms
The moments of inertia of the circles are thus
- Circle 1 moment of inertia: m1 * r1^2 = ( 152.46 kg)(
6.6 meters) ^ 2 = 1006.236 kg m ^ 2,
- Circle 2 moment of inertia: m2 * r2^2 = ( 304.92 kg)(
13.2 meters) ^ 2 = 2012.472 kg m ^ 2, and
- Circle 3 moment of inertia: m3 * r3^2 = ( 457.38 kg)(
19.8 meters) ^ 2 = 3018.708 kg m ^ 2.
The total moment of inertia is therefore the sum
6037.416 kg m^2 of these mr ^ 2 contributions.
A torque of 1235 meter Newtons will thus result in an
acceleration of
- angular acceleration = `alpha = `tau / I = 1235 meter
Newtons / 6037.416 kg m ^ 2 = .2045577 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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