This worksheet is required to provide values to be used in f

This worksheet is required to provide values to be used in future labs. The following information is needed to complete the assignment. From manufacture specifications, motor inertia, J motor = 3.8 times 10^-5 Kg-m^2. (Note for the future: Motor to drive disk gear ratio = 1). Thickness of drive disk plate = 0.47 cm. Diameter of drive disk = 13.21 cm. Plate material density, p aluminum = 2.71 g/cm^3. (Neglect cutout slots). The .500Kg weight diameter = 4.95cm. The .200Kg weight diameter = 3.1 cm. Find the following inertia (J) amounts in Kg-m^2. J_Drive Disk = J_w(.5 kg) = J_w(.2 kg) = Now, find the following configurations; Inertia Configurations;

Solution

Answer for the first part - Inertia of the parts

1) Inertia of Drive Disk

Drive disk can be assumed to be a cylinder rotating about its own axis,

Moment of inertia J₁ = Mr² / 2 where M is the mass of the disk and r is the radius of the disk

Mass M = Density * Volume ;

Volume = area * height;

area = pi * d² / 4 = 3.14 * (0.1321)² / 4 = 0.0137m²

height = thickness of disk = 0.47cm = 0.0047m , Volume = 0.0137 * 0.0047 = 6.438 * 10^-5 m³

Mass M = Density * Volume ;

Density = 2.71g/cm³ = 2.71 * 10³ Kg / m³   

M = 2.71 * 10³ * 6.438 * 10^-5 = 0.1745kg

Mass = 0.1745kg

Radfius = r = 0.1321/2 = 0.066m ;

Moment of inertia J₁ = 0.1745 * 0.066² / 2 = 3.8 * 10^-5 Kgm^2;

  J₁ = 3.8 * 10^-5 Kgm^2;

2) Inertia of the two weights

Moment of inertia J = mass * radius²

J₂ = M₂ * R₂² = 0.5 * (4.95 / (2 * 100) ) ² = 0.5 * 0.02475² = 3.063 * 10^-4

J₂ = 3.063 * 10^-4 Kgm²

J3 = M3 * R₃² = 0.2 * ( 3.1 / (2 * 100 ) ) ² = 4.805 * 10^-5

J₃ = 4.805 * 10^-5Kgm²