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 5.5 kilograms and the other of mass 9.7 kilograms. The system is constrained to rotate about an axis of rotation. The first object is 6.36 meters and the second 14 meters from the axis of rotation. A torque of 16 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 5.5 kilograms at a constant distance of 6.36 meters from the axis of rotation, and it contributes mr ^ 2 = ( 5.5 kg)( 6.36 m) ^ 2 = 222.4728 kg m ^ 2 to the moment of inertia.
• The second mass consists of 9.7 kilograms at a constant distance of 14 meters from the axis of rotation, and it contributes mr ^ 2 = ( 9.7 kg)( 14 m) ^ 2 = 1901.2 kg m ^ 2 to the moment of inertia.

• The moment of inertia of the system is therefore I = `Sigma mr ^ 2 = 222.4728 kg m ^ 2 + 1901.2 kg m ^ 2 = 2123.673 kg m^2.

The acceleration resulting from a torque of 16 meter Newtons will therefore be `alpha = `tau /( `Sigma mr ^ 2) = ( 16 meter Newtons) / ( 2123.673 kg m^2) = 7.534117E-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