Mass Moment of Inertia
Title: Mass Moment of Inertia
Objective:
To determine mass moment of inertia of a part using experimental method.
Theory:
If a part has been designed and built, its mass moment of inertia can be determined approximately by a simple experiment. This requires that the part be swung about any axis (other than one that passes through its CG) parallel to that about which the moment is sought and its period of pendular oscillation measured. Figure 1 shows a part of connecting rod suspended on a knife-edge pivot at ZZ and rotated through a small angle θ.
Its weight force W acts as its CG has a component W sin θ perpendicular to the radius r from the pivot to the CG.
From rotational form of Newton’s equation:
TZZ=IZZ∝
Substituting equivalent expressions for TZZ and ∝;
-Wsin θr=IZZd2θdt2
Where the negative sign is used because the torque is in the opposite direction to angle a.
For small values of θ, sinθ=θ, approximately, so:
-Wθr=IZZd2θdt2
d2θdt2=-WrIZZθ
Equation above is a second order differential equation with constant coefficients that has the well-known solution: θ=CsinWrIZZt+DcosWrIZZt The constants of integration C and D can be found from the initial conditions defined at the instant the part of released and allowed to swing.
At: t=0, θ=θmax, ω=dθdt=0;then:C=0, D=θmax
And:
θ=θmaxcosWrIZZt
Equation above defines the part’s motion as a cosine wave that completes a full cycle of period τ sec when
WrIZZτ=2π
The weight of the part is easily measured. The CG location can be found be balancing part on a knife edge or suspending it from two different locations, either approached giving the distance r. The period of oscillation τ can be measured with a stopwatch, and averaging over a number of cycles to reduce experimental error. With these data, equation above can be solved for the mass moment of inertia IZZ about the pivot ZZ at:
IZZ=Wrτ2π2
And the moment of inertia IGG about the CG can then be found using the transfer theorem:
IZZ=IGG+mr2
Objective:
To determine mass moment of inertia of a part using experimental method.
Theory:
If a part has been designed and built, its mass moment of inertia can be determined approximately by a simple experiment. This requires that the part be swung about any axis (other than one that passes through its CG) parallel to that about which the moment is sought and its period of pendular oscillation measured. Figure 1 shows a part of connecting rod suspended on a knife-edge pivot at ZZ and rotated through a small angle θ.
Its weight force W acts as its CG has a component W sin θ perpendicular to the radius r from the pivot to the CG.
From rotational form of Newton’s equation:
TZZ=IZZ∝
Substituting equivalent expressions for TZZ and ∝;
-Wsin θr=IZZd2θdt2
Where the negative sign is used because the torque is in the opposite direction to angle a.
For small values of θ, sinθ=θ, approximately, so:
-Wθr=IZZd2θdt2
d2θdt2=-WrIZZθ
Equation above is a second order differential equation with constant coefficients that has the well-known solution: θ=CsinWrIZZt+DcosWrIZZt The constants of integration C and D can be found from the initial conditions defined at the instant the part of released and allowed to swing.
At: t=0, θ=θmax, ω=dθdt=0;then:C=0, D=θmax
And:
θ=θmaxcosWrIZZt
Equation above defines the part’s motion as a cosine wave that completes a full cycle of period τ sec when
WrIZZτ=2π
The weight of the part is easily measured. The CG location can be found be balancing part on a knife edge or suspending it from two different locations, either approached giving the distance r. The period of oscillation τ can be measured with a stopwatch, and averaging over a number of cycles to reduce experimental error. With these data, equation above can be solved for the mass moment of inertia IZZ about the pivot ZZ at:
IZZ=Wrτ2π2
And the moment of inertia IGG about the CG can then be found using the transfer theorem:
IZZ=IGG+mr2