WAFring1ВEľcentral72°DPA prototype wheel cover consists of a thin ring, five rectangular plates, and a central circular plate.Given:Each plate has a mass of: m_plate = 0.8 kgThe ring has a mass of: m_ring = 1.4 kg .The ring has a radius of: r_ring= 30cmThe central plate has radius of: r_central= 9cmEach rectangular plate has a length I= 21 cm and width w = 6 cmEach plate is spaced 72 ° apart from one anotherFind: What is the mass moment of inertia if the wheel cover rotates about point P? Assume the thickness of the ring is negligible.

Answered: A Iring 1 В E G Icentral 72° D P A…

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

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W
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Fring
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72°
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A prototype wheel cover consists of a thin ring, five rectangular plates, and a central circular plate.
Given:
Each plate has a mass of: m_plate = 0.8 kg
The ring has a mass of: m_ring = 1.4 kg .
The ring has a radius of: r_ring= 30cm
The central plate has radius of: r_central= 9cm
Each rectangular plate has a length I= 21 cm and width w = 6 cm
Each plate is spaced 72 ° apart from one another
Find: What is the mass moment of inertia if the wheel cover rotates about point P? Assume the thickness of the ring is negligible.

Expert Solution

Step 1 Calculate mass moment of inertia

Assume that the mass moment of inertia of central part of ring will be $I_{I R} =$ 0 ; due to there is not given the mass of central plate. but assume that mass of central late is 0.1 kg so

Mass moment of inertia of central plate is = $0.1 \times 9^{2} = 8.1 K g . c m^{2};$

Mass moment of inertia due to rectangular plate will be = $I_{s p o k e} = \frac{1}{12} [m a s s o f p l a t e \times (l^{2} + w^{2})] = \frac{1}{12} [0.8 \times (24^{2} + 6^{2})] = 40.8 K g . c m^{2};$

So the total no of spokes=5

The total moment of inertia due to spokes= $5 \times I_{s p o k e} = 5 \times 40.8 = 204 K g . c m^{2};$

Mass moment of inertia of outer ring will be = $I_{O R} = 1.4 \times 30^{2} = 1260 K g . c m^{2};$

So the total mass moment of inertia about G will be $I_{G}$ =8.1+204+1260=1472.1 $K g . c m^{2}$ ;

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