Set 8 Problem number 13

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s); Extension

• Figure

Problem

A system consists of two concentric hoops constrained to rotate about an axis through their centers and perpendicular to their common plane. A torque of 23.5 meter Newtons is applied to the system.

The first hoop has radius 6.48 meters and mass .3 kilograms, while the second has radius 17 meters and mass 9.8 kilograms.

• What is the moment of inertia of the system and what will be its angular acceleration?

Solution

Each mass rotates at a constant distance from the center of rotation.

• The first mass consists of .3 kilograms at a constant distance of 6.48 meters from the axis of rotation, and it contributes mr ^ 2 = ( .3 kg)( 6.48 m) ^ 2 = 12.59712 kg m ^ 2 to the moment of inertia.
• The second mass consists of 9.8 kilograms at a constant distance of 17 meters from the axis of rotation, and it contributes mr ^ 2 = ( 9.8 kg)( 17 m) ^ 2 = 2832.2 kg m ^ 2 to the moment of inertia.

• The moment of inertia of the system is therefore I = `Sigma mr ^ 2 = 12.59712 kg m ^ 2 + 2832.2 kg m ^ 2 = 2844.797 kg m^2.

The acceleration resulting from a torque of 23.5 meter Newtons will therefore be `alpha = `tau /( `Sigma mr ^ 2) = ( 23.5 meter Newtons) / ( 2844.797 kg m^2) = 8.260694E-03 rad/s ^ 2.

Generalized Solution

If we have point masses m1, m2, ..., mn along a rod at distances r1, r2, ..., rn from the center of rotation, then we have individual moments of inertia m1 r1^2, m2 r2^2, ..., mn rn^2.

• The total moment of inertia will therefore be I = `sigma (m r^2) = m1 r1^2 `m2 r2^2 0... `mn rn^2.
• The angular acceleration resulting from a net torque `tau will therefore be

• angular acceleration = `tau / I = `tau / [ `sigma (m r^2)] = `tau / [ m1 r1^2 `m2 r2^2 0... `mn rn^2 ].

Explanation in terms of Figure(s); Extension

The figure below shows two masses m1 and m2 along a massless rod which constrains them to rotate about a central axis at respective distances r1 and r2 from the axis.

• Their individual moments of inertia are I1 = m1 r1^2 and I2 = m2 r2^2.
• The total moment of inertia is therefore I = I1 `I2 = m1 r1^2 `m2 r2^2.
• The angular acceleration resulting from an applied torque `tau is therefore

• Angular acceleration = `alpha = `tau / I = `tau / (m1 r1^2 `m2 r2^2).

Figure