EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
9th Edition
ISBN: 8220100663987
Author: Jewett
Publisher: Cengage Learning US
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Chapter 20, Problem 20.65AP

(a)

To determine

The change in kinetic energy of the disk.

(a)

Expert Solution
Check Mark

Answer to Problem 20.65AP

The change in kinetic energy of the disk is 9.31×1010J .

Explanation of Solution

Given info: The radius of copper disk is 28.0m , thickness of copper disk is 1.20m , the temperature at collision is 850°C , angular speed of copper disk is 25.0rad/s and the temperature after radiation is 20°C .

Write the equation for change in kinetic energy of the disk.

ΔK.E.=12Ifωf212Iiωi2=12(Ifωf)ωf12Iiωi2

Here,

Ii is the initial moment of inertia of the disk.

If is the final moment of inertia of the disk.

ωi is the initial angular speed of the disk.

ωf is the final angular speed of the disk.

Write the equation of conservation of angular momentum.

Iiωi=Ifωf

Substitute Iiωi for Ifωf in the above equation to get the change in kinetic energy of the disk.

ΔK.E.=12(Ifωf)ωf12Iiωi2=12(Iiωi)ωf12Iiωi2=12Iiωi(ωfωi) (1)

Write the formula for initial moment of inertia Ii .

Ii=12mr2=12(ρV)r2=12(ρ(πr2t))r2=12ρπr4t

Here,

m is the mass of the disk.

r is the radius of disk.

ρ is the density of copper disk.

t is the thickness of copper disk.

The density of copper is 8920kg/m3 .

Substitute 8920kg/m3 for ρ , 28.0m for r and 1.20m for t in the above equation to get the initial moment of inertia.

Ii=12(8920kg/m3)π(28.0m)4(1.20m)=1.033×1010kgm2

Write the equation of conservation of angular momentum to calculate the final angular speed of the disk.

Iiωi=Ifωf12mri2ωi=12mrf2ωf12mri2ωi=12mri2(1α|ΔT|)2ωfωf=ωi(1α|ΔT|)2

Further solve the above equation to calculate the final angular speed of the disk.

ωf=ωi(1α|ΔT|)2=ωi(1α(TcTr))2

Here,

α is the coefficient of linear expansion for copper.

ΔT is the change in temperature.

Tc is the temperature at collision.

Tr is the temperature after radiation.

The value of coefficient of linear expansion α for copper is 17×106°C1 .

Substitute 25.0rad/s for ωi , 17×106°C1 for α , 850°C for Tc and 20°C for Tr in the above equation to get the final angular speed of the disk.

ωf=(25.0rad/s)(1(17×106°C1)(850°C20°C))2=25.7207rad/s

Substitute (1.033×1010kgm2) for Ii , 25.0rad/s for ωi and 25.7207rad/s for ωf in equation (1) to calculate the change in kinetic energy of the disk.

ΔK.E.=12(1.033×1010kgm2)(25.0rad/s)((25.7207rad/s)(25.0rad/s))=9.31×1010J

Conclusion:

Therefore, the change in kinetic energy of the disk is 9.31×1010J .

(b)

To determine

The change in internal energy of the disk.

(b)

Expert Solution
Check Mark

Answer to Problem 20.65AP

The change in internal energy of the disk is 8.47×1012J .

Explanation of Solution

Given info: The radius of copper disk is 28.0m , thickness of copper disk is 1.20m , the temperature at collision is 850°C , angular speed of copper disk is 25.0rad/s and the temperature after radiation is 20°C .

Write the equation to calculate the change in internal energy of the disk.

ΔEint=Q=mcΔT=(ρV)c(TrTc)=ρπr2tc(TrTc)

Here,

ΔEint is the change in internal energy of the disk.

Q is the energy required to change the temperature of substance.

c is the specific heat of the copper disk.

Specific heat of copper disk is 387J/kg°C .

Substitute 8920kg/m3 for ρ , 28.0m for r, 1.20m for t, 387J/kg°C for c, 20°C for Tr and 850°C for Tc in the above equation to get the change in internal energy of the disk.

ΔEint=(8920kg/m3)π(28.0m)2(1.20m)(387J/kg°C)(20°C850°C)=8.4683×1012J8.47×1012J

Conclusion:

Therefore, the change in internal energy of the disk is 8.47×1012J .

(c)

To determine

The amount of radiated energy.

(c)

Expert Solution
Check Mark

Answer to Problem 20.65AP

The amount of radiated energy is 8.38×1012J .

Explanation of Solution

Given info: The radius of copper disk is 28.0m , thickness of copper disk is 1.20m , the temperature at collision is 850°C , angular speed of copper disk is 25.0rad/s and the temperature after radiation is 20°C .

Write the equation for change in kinetic energy of the disk to calculate the amount of radiated energy.

ΔK.E.=TERΔEintTER=ΔK.E.+ΔEint

Here,

TER is the amount of radiated energy.

Substitute 9.31×1010J for ΔK.E. and 8.47×1012J for ΔEint in the above equation to get the amount of radiated energy.

TER=(9.31×1010J)+(8.47×1012J)=(9.31×1010J)(8.47×1012J)=8.38×1012J

Conclusion:

Therefore, the amount of radiated energy is 8.38×1012J .

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Chapter 20 Solutions

EBK PHYSICS FOR SCIENTISTS AND ENGINEER

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