Question

Two cylindrical iron rods both at temperatureThave lengthsLand 2L, and radiiRandR/2, respectively. If the rods’ temperatures are raised to 3T, and forces are appliedto prevent the rods from changing length, what is the ratio of the forces on the tworods?

Answer 1

For 1st rod -

Initial temperature = T

Final temperature = 3T

Change in Temperature, = 3T - T = 2T

Length = L

Radius of cross section = R

For 2nd rod -

Initial temperature = T

Final temperature = 3T

Change in Temperature, = 3T - T = 2T

Length = 2L

Radius of cross section = R/2

This problem will be solved by referring to modulus of elasticity of longitudinal objects i.e. Young's Modulus (Y)

We know that,

where, Stress = F/A i.e. Force F acting on area of cross section A of the material to produce a deformation in the length of the object. Therefore, we can also say, that we have to apply equal and opposite Force = F to prevent the rods from changing it's length. ( This is the Force F, we have to find in this question )

is the strain or deformity produced in the object of original length L

But, in this question, the strain or deformity in the object has originated from heating up of the object, therefore, thermal expansion has taken place, in which rod has linearly expanded from it's original shape.

This Thermal Expansion is given by :-

where is defined as strain as we have already discussed

is the coefficient of the linear thermal expansion

is the change in Temperature

Now substituting eqn.(2) in eqn.(1) :-

Now, For 1st rod :-

For 2nd rod :-

Notice the Y and are same for both the rods, because these two quantities depend on the material of which the object is made up of, and not on the dimensions of the object. Since both the rods are made of Iron, they will have the same value for both rods.

Therefore, eqn.(4) and (5) are equal. Equating them, we get :-

  ( and cancel out )