This worksheet is required to provide values to be used in future labs. The foll

Question

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;

Explanation / Answer

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 Kgm2;

  J1 = 3.8 * 10-5 Kgm2;

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

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