VARIATION PROBLEMS:

In your Algebra class, you discussed the two types of variation: a direct proportion and an inverse proportion. A third type is when a variable say, z, varies directly with the product of two variables x and y,  that is

Where k is the constant of proportionality between z and the product xy. The variable z is said to vary jointly with x and y.  One good example of joint variation is momentum. The momentum of an object  varies directly with  the product of its mass and  velocity (k = 1).

 

EXAMPLE 1: 

The kinetic energy of a body KE  is given by the equation

Where m represents the mass of the object and v, its velocity.  The equation, tells us that KE varies directly with the product of  m and v2 where the constant of proportionality k = 1/2. Suppose  a car  of mass mA, acquires a  kinetic energy  KEA of  10,000 J when it moves at  a velocity vA . What will be the kinetic energy of a second car that  is twice as heavy and twice as fast as the first car?

SOLUTION: 

                The equation above can also be written as (following the technique suggested in “USING PROPORTION”),

This suggests that,

 

Solving for KEB,

Substituting   and  ,

 

In many instances, direct, inverse and joint variation are combined. We then have a compound variation. A good example of such a compound variation is the dependence of  centripetal force on the mass, velocity and radius of the circle of revolution.  The centripetal force is given by the expression

 Equation (1)

Or the centripetal force varies directly with the mass, with the square of the velocity, and it varies inversely with the radius of revolution. Since the velocity v is equal to  , where T is the period of revolution, we can write the Equation (1)  as

 Equation (2)

Equation (2) shows that for a given body, the centripetal force varies directly with the radius of revolution and inversely with the square of the period. The question that arises is why the centripetal force  varies inversely with R in Equation (1)  but that  varies directly with R in Equation (2). We should, however, bear in mind  what other quantities  are implied as being constant. In (1) ,   for a given body moving with constant speed, while in (2) ,  for a body whose period of rotation is kept constant.

 

EXAMPLE 2:

                A stone is whirled in a horizontal circle  by means of a string  50 cm long and with a period of 0.50 s. If the tension in the string is 20 N, what is the tension if the string is shortened to 25 cm and the period is also shortened to 0.25 s? (Note: the tension in the string provides the centripetal force for the stone to follow a circular trajectory)

 

SOLUTION:

It is implied in the problem that

Where 

Rearranging the equation so that the constant k is on one side  and the variables  are on  the other ,

(equation A)

If we let  the new tension new radius  and the new period be , , respectively, Equation A, gives us

Solving for  ,

 

EXAMPLE 3:

                Hooke’s Law states that within the elastic limit, the stress   is proportional to strain . The constant of proportionality is known as Young’s modulus of elasticity, Y. In equation

A steel wire of length L, of cross-sectional area A is subjected to a tension T. The elongation is observed to be 1 mm. What will be the elongation of another steel wire of length 2L, of cross sectional area ½ A, if the tension is 2T?

 

SOLUTION:

Since the second wire is made of steel, Y for the second wire is the same as that for the first.  That is

Solving for  ,