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A moment M is slowly increased as it is applied to the disk shown in Fig. tential energy of the system with respect to a reference at e = 0° as the disk is rotated to the position 0 = 45°. The spring (k = 10 lb/in.) is unstretched at e = 0°. Determine the change in po- W = 50lb 16 in. M 8 in. www A

A moment M is slowly increased as it is applied to the disk shown in Fig. tential energy of the system with respect to a reference at e = 0° as the disk is rotated to the position 0 = 45°. The spring (k = 10 lb/in.) is unstretched at e = 0°. Determine the change in po- W = 50lb 16 in. M 8 in. www A

Elements Of Electromagnetics
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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A moment M is slowly increased as it is applied to the disk shown in Fig. tential energy of the system with respect to a reference at e = 0° as the disk is rotated to the position 0 = 45°. The spring (k = 10 lb/in.) is unstretched at e = 0°. Determine the change in po- W = 50lb 16 in. M 8 in. www A
Transcribed Image Text:A moment M is slowly increased as it is applied to the disk shown in Fig. tential energy of the system with respect to a reference at e = 0° as the disk is rotated to the position 0 = 45°. The spring (k = 10 lb/in.) is unstretched at e = 0°. Determine the change in po- W = 50lb 16 in. M 8 in. www A
Expert Solution
Step 1

Given Data:

  • The spring constant is k = 10 lb/in.
  • The angle of rotation is θ=45∘.
  • The radius of the disk is r = 8 in.
  • The distance of weight W from the center is D = 16 in.
  • The given value of weight is W= 50 lb

The configuration diagram can be drawn as,

Mechanical Engineering homework question answer, step 1, image 1

 

 

 

 

 

 

Step 2

The stretch in the spring (δ) can be determined as,

δ=AA'=rθ=8 in×45∘×π rad180∘=6.28 in

From the geometry OP can be determined as,

cos45∘=OPOROP=ORcos45∘OP=16cos45∘ in=11.31 in

The fall in height (h) of the weight W can be determined as,

h=QP=OQ-OP=16 in-11.31in=4.69 in

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