Set 8 Problem number 14

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s); Extension

• Figure

Problem

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 7.1 meters, the second twice and the third three times that radius. Around each circle, masses of 10 kilograms are evenly distributed, with 1.42 / `pi meters of arc between masses.

• A net torque of 1062 meter Newtons is applied. What angular acceleration results?

Solution

The first circle has a total circumference of 2 `pi ( 7.1 meters).

• With masses spaced at `pi / 1.42 meters, the number of masses on the first circle will be

• number of masses on first circle = circumference / arc distance between masses = 2 `pi ( 7.1 meters)/( `pi / 1.42 meters) = 2( 7.1 )( 1.42 ) = 20.164 .

• 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( 20.164 ) and 3( 20.164 ) masses on the second third circles.

At 10 kilograms per mass, the circles will thus have masses

• Circle 1 total mass: m1 = 201.64 kilograms
• Circle 2 total mass: m2 = 403.28 kilograms
• Circle 3 total mass: m3 = 604.92 kilograms

The moments of inertia of the circles are thus

• Circle 1 moment of inertia: m1 * r1^2 = ( 201.64 kg)( 7.1 meters) ^ 2 = 1431.644 kg m ^ 2,
• Circle 2 moment of inertia: m2 * r2^2 = ( 403.28 kg)( 14.2 meters) ^ 2 = 2863.288 kg m ^ 2, and
• Circle 3 moment of inertia: m3 * r3^2 = ( 604.92 kg)( 21.3 meters) ^ 2 = 4294.932 kg m ^ 2.

The total moment of inertia is therefore the sum 8589.863 kg m^2 of these mr ^ 2 contributions.

A torque of 1062 meter Newtons will thus result in an acceleration of

• angular acceleration = `alpha = `tau / I = 1062 meter Newtons / 8589.863 kg m ^ 2 = .1236341 rad/second ^ 2.

Generalized Solution

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).

Explanation in terms of Figure(s); Extension

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.

Figure