Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 13, Problem 107P

(a)

To determine

The temperature must the ring be heated to fit over the shaft.

(a)

Expert Solution
Check Mark

Answer to Problem 107P

The temperature must the ring be heated to fit over the shaft is 44°C_.

Explanation of Solution

Write the expression for change in length

ΔL=αL0ΔT (I)

Here, ΔL is the change in length, α is the coefficient of linear expansion, ΔT is the change in temperature

Replace, ΔT by TT0, and rearrange the equation (I) to obtain an expression for T

ΔL=αL0(TT0)T=ΔLαL0+T0 (II)

Conclusion:

Substitute 0.00200cm for ΔL, 12×106K1 for α, 7.00000cm for L0, 20.0°C for T0 in equation (II)

T=0.00200cm12×106K1×7.00000cm+20.0°C=43.8°C44°C

Therefore, the temperature must the ring be heated to fit over the shaft is 44°C_.

(b)

To determine

The temperature up to which shaft and ring must be cooled to allow the ring to slide off the shaft again.

(b)

Expert Solution
Check Mark

Answer to Problem 107P

The temperature up to which shaft and ring must be cooled to allow the ring to slide off the shaft again 21°C_

Explanation of Solution

Consider the diameters of shaft and ring are equal

Ls=Lb (I)

Here, Ls is the diameter of steel ring, and Lb is the diameter of brass shaft

According to the expression for fractional change in length Ls can be written as

ΔLs=αsLs0ΔTLsLs0=αsLs0ΔTLs=Ls0+αsLs0ΔT (II)

Here, Ls0 is the diameter at absolute temperature, αs is the linear explanation coefficient od steel ring, and ΔT is the change in temperature

Similarly in case of Lb

Lb=Lb0+αbLb0ΔT (III)

Here, Lb0 is the diameter of shaft at absolute temperature, αb0 is the linear expansion coefficient of shaft, and ΔT is the temperature change

Equate equation (II) and (III) replace ΔT by TT0 and rearrange to obtain an expression for T

Ls0+αsLs0ΔT=Lb0+αbLb0ΔT(TT0)(αsLs0αbLb0)=Lb0Ls0T=T0+Lb0Ls0(αsLs0αbLb0) (IV)

Conclusion:

Substitute 20.0°C for T0, 7.00200cm for Ls0, 7.00000cm  for Lb0, 12×106K1 for αs0, 19×106K1 for αb0 in equation (IV)

T=20.0°C+7.00200cm7.00000cm (12×106K1×7.00000cm19×106K1×7.00200cm)=21°C

Therefore, the temperature up to which shaft and ring must be cooled to allow the ring to slide off the shaft again 21°C_.

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Chapter 13 Solutions

Physics

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