Set 8 Problem number 15

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s); Extension

• Figure

Problem

Find the angular acceleration of the system if a torque of 1958 meter Newtons is applied to a system consisting of three concentric hoops constrained to rotate together about an axis through their common center and perpendicular to their common plane.   Each hoop has a mass density of 2.228164 kilograms/meter, and the hoop radii are 3.5 meters, 7 meters and 10.5 meters.

Solution

The first circle or hoop has circumference 2 `pi ( 3.5 meters) = 21.98 meters.

• Its mass is therefore 21.98 meters ( 2.228164 kilograms/meter) = 48.97505 kg.

The second circle or hoop has circumference 2 `pi ( 7 meters) = 43.96 meters.

• Its mass is therefore 43.96 meters ( 2.228164 kilograms/meter) = 97.9501 kg.

The third circle or hoop has circumference 2 `pi ( 10.5 meters) = 65.94 meters.

• Its mass is therefore 65.94 meters ( 2.228164 kilograms/meter) = 146.9251 kg.

The three moments of inertia will therefore be

• hoop 1: ( 48.97505 kilograms)( 3.5 meters) ^ 2 = 1799.833 kg m ^ 2,
• hoop 2: ( 97.9501 kilograms)( 7 meters) ^ 2 = 4799.555 kg m ^ 2, and
• hoop 3: ( 146.9251 kilograms)( 10.5 meters) ^ 2 = 5399.499 kg m ^ 2.

These add up to the total moment of inertia

• total moment of inertia = I = 11998.89 kg m ^ 2.

The 1958 meter Newton torque will result in an angular acceleration of `alpha = `tau / I = 1958 meter Newtons / ( 11998.89 kilogram meter ^ 2) = .1631818 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) ].

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 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.

Figure