Question

The roller assembly, shown in figures 3, is used as part of a machine in a manufacturing line.

- Determine the radius of gyration of part A.Data:Each part A has mass m subscript A equals 110.3 gEach part B has mass m subscript B equals 50 over 3 m subscript AEach part C has mass m subscript C equals 8 over 3 m subscript BPart D has m subscript D equals 117 over 40 m subscript C

- Determine the mass of part B.Data:Each part A has mass m subscript A equals 110.3 gEach part B has mass m subscript B equals 50 over 3 m subscript AEach part C has mass m subscript C equals 8 over 3 m subscript BPart D has m subscript D equals 117 over 40 m subscript C

- Determine the moment of inertia of part C.Data:Each part A has mass m subscript A equals 110.3 gEach part B has mass m subscript B equals 50 over 3 m subscript AEach part C has mass m subscript C equals 8 over 3 m subscript BPart D has m subscript D equals 117 over 40 m subscript C

C B 15 mmT A Figures 3 200 mm 190 mm 100 mm

Answer 1

Each part A,B,C,D is a cylinder.

The part A has a diameter of 15 mm.

Now, the radius of gyration is assumed to be the imaginary point , distant k from the axis of rotation, that has a mass equal to the object. It has the property that

Iimg Mk Iobi

The for part A is the moment of inertia of a solid cylinder along its axis of symmetry. It is given by:

ob 1

obj Mr 2.

On equating these, we get:

Mr2 Mk2 k 2 10.606mm V2

This is the radius of gyration of the parts labelled A.

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The math part of the question is highly difficult to read. I will thus assume this for the second part:

117 me 40 50 8 mb md 3 ma 110.3g; mb=ma me 3

From this, we can easily see that

50 110.3g 1833.3g 3

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The part C is also a cylinder of diameter 200 mm = 0.2 m => Radius =0.1 m

Its mass is

1833.3g4888.8g

The moment of inertia is:

kg m^2

x 4.89kg x 0.12m2 = 0.02445 -mr l