Set 8 Problem number 13

• Important Note on Notation

• Problem

• Solution

• Generalized Solution

• Explanation in terms of Figure(s); Extension

• Figure

Problem

A massless disk holds two objects, one of mass 9.900001 kilograms and the other of mass 9 kilograms. The system is constrained to rotate about an axis of rotation. The first object is 5.12 meters and the second 16 meters from the axis of rotation. A torque of 8.25 meter Newtons is applied to the system.

• 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 9.900001 kilograms at a constant distance of 5.12 meters from the axis of rotation, and it contributes mr ^ 2 = ( 9.900001 kg)( 5.12 m) ^ 2 = 259.5226 kg m ^ 2 to the moment of inertia.
• The second mass consists of 9 kilograms at a constant distance of 16 meters from the axis of rotation, and it contributes mr ^ 2 = ( 9 kg)( 16 m) ^ 2 = 2304 kg m ^ 2 to the moment of inertia.

• The moment of inertia of the system is therefore I = `Sigma mr ^ 2 = 259.5226 kg m ^ 2 + 2304 kg m ^ 2 = 2563.522 kg m^2.

The acceleration resulting from a torque of 8.25 meter Newtons will therefore be `alpha = `tau /( `Sigma mr ^ 2) = ( 8.25 meter Newtons) / ( 2563.522 kg m^2) = 3.218228E-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