Problems in Mathematical Analysis.pdf - pwp.net.ipl.pt

174 Definite Integrals [Ch. 5
and
Solution. We have:
10
p t*
.-Jo.irt-o.i T = 250 metres
=-=25 m/sec.
2. The work of a force. If a variable force X=f(x) acts in the direction
of the x-axis, then the work of this force over an interval [x ly x z ] is
A =
Example 2. What work has to be performed to stretch a spring 6 cm, if
a force of 1 kgf stretches it 1 by cm?
Solution, According to Hook's law the force X kgf stretching the spring
by xm is equal to X = kx, where k is a proportionality constant.
Putting x = 0.01 m and X = l kgf, we get = 100 and, hence, X = 100v.
Whence the sought-for work is
0.06 0.08
A = j 100 x dx = 50 x 2 = 0. 18 kgm
3. Kinetic energy. The kinetic energy of a material point of mass m and
velocity v is defined as
mv*
The kinetic energy of a system of n material points with masses
mv m 2% ..., m n having respective velocities t; lf v 2 , ..., v n , is equal to
To compute the kinetic energy of a solid, the latter is appropriately partitioned
into elementary particles (which play the part of material points); then
by summing the kinetic energies of these particles we get, in the limit, an
integral in place of the sum (1).
Example 3. Find the kinetic energy of a homogeneous circular cylinder
of density 6 with base radius R and altitude h rotating about its axis with
angular velocity CD.
Solution. For the elementary mass dm we take the mass of a hollow
cylinder of altitude h with inner radius r and wall thickness dr (Fig. 60).
We have:
Since the linear velocity of the mass dm is equal to t; = /-co, the elementary
kinetic energy is