Vector Mechanics For Engineers
Vector Mechanics For Engineers
12th Edition
ISBN: 9781259977305
Author: BEER, Ferdinand P. (ferdinand Pierre), Johnston, E. Russell (elwood Russell), Cornwell, Phillip J., SELF, Brian P.
Publisher: Mcgraw-hill Education,
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Chapter 19.3, Problem 19.98P
To determine

(a)

The period of vibration of the shell when it is displaced vertically and released.

Expert Solution
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Answer to Problem 19.98P

The period of vibration: τn=0.352 s

Explanation of Solution

Given information:

Mass of the hollow-spherical shell is 500-g.

Radius of the spherical shell is 80 mm.

Constant of the spring is 500 N/m.

Calculations:

Vector Mechanics For Engineers, Chapter 19.3, Problem 19.98P

For the two positions as shown in the figure:

At Position 2:Kinetic energy: T2=0Potential energy: V2=12kxm2

At Position 1 :Kinetic energy: T1=Tspere+Tfluid=12msvm2+14ρVvm2Potential energy: V1=0Now, applying principles of conservation of energy and simple harmonic motion.T1+V1=T2+V2:12msvm2+14ρVvm2+0=0+12kxm2(vm=x˙m=xmωn)12(ms+14ρV)xm2ωm2=12kxm2ωn2=kms+12ρVsubstituting values,ωn2=500 N/m(0.5 kg)+(12ρV)12ρV=12(1000 kg/m3)(43π(0.08 m)3)12ρV=1.0723kgωn2=500 N/m(0.5 kg)+(1.0723kg)=318s¯2Period of vibration.τn=2πωn=2π318τn=0.352 s

Conclusion:

The period of vibration of the shell when it is displaced vertically and released is τn=0.352 s.

To determine

(b)

The period of vibration of the shell if the tank is accelerated upward at the constant rate of 8 m/s2.

Expert Solution
Check Mark

Answer to Problem 19.98P

The period of vibration: τn=0.352 s

Explanation of Solution

Given information:

Mass of the hollow-spherical shell is 500-g.

Radius of the spherical shell is 80 mm.

Constant of the spring is 500 N/m.

Calculations:

Accelerating the tank vertically upward does not change the massHence the period of vibration remains same as calculated in part (a).

For the two positions as shown in the figure:

At Position 2:Kinetic energy: T2=0Potential energy: V2=12kxm2

At Position 1 :Kinetic energy: T1=Tspere+Tfluid=12msvm2+14ρVvm2Potential energy: V1=0Now, applying principles of conservation of energy and simple harmonic motion.T1+V1=T2+V2:12msvm2+14ρVvm2+0=0+12kxm2(vm=x˙m=xmωn)12(ms+14ρV)xm2ωm2=12kxm2ωn2=kms+12ρVsubstituting values,ωn2=500 N/m(0.5 kg)+(12ρV)12ρV=12(1000 kg/m3)(43π(0.08 m)3)12ρV=1.0723kgωn2=500 N/m(0.5 kg)+(1.0723kg)=318s¯2Period of vibration.τn=2πωn=2π318τn=0.352 s

The period of vibration of the shell when it is displaced vertically and released is τn=0.352 s.

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Chapter 19 Solutions

Vector Mechanics For Engineers

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