# 16. Advance Illustration | Rigid Body Dynamics | A Rotating Tilted Rod | by Ashish Arora

in this illustration, we’ll be analyzing,
a rotating tilted rod. here the figure shows a thin uniform rod ay b of mass m and length
l rotating about the vertical axis o o dash. at angular velocity omega. here we are required
to find the torque of centrifugal force is acting on rod about the midpoint, of the rod
c. in the frame of reference rotating at angular speed omega, attached to the axis o o dash.
so in this situation here to analyze, the centrifugal forces which act in the rotating
frame of reference here, we can redraw the figure of rod, and analyze on it. we can see
different points, of the rod are at different distances from the axis of rotation o o-dash.
so, we consider an element at a distance x. from the center and it is of width, d x. for
this element if we consider its mass. we can write. mass. of an element. taken. as shown,
is. this we can write as. m by l multiplied by d x. and this element is considered to
be, revolving in a circle of radius if this angle is theta, then its radius r can be written
as, x sine theta. so in this situation we can say. the centrifugal force. on element.
due to, its, circular motion, is. the centrifugal force d f we can write, which is acting in
outward direction and the value of d f we can write as d m. omega square, r. the value
of r we can substitute as x sine theta so this is d m omega square, x sine theta. now
if we calculate the torque, due to d f. on rod. about, c is. this can be written as d
tau and that will be equal to d f. multiplied by this distance here we can see from point
c. the line of action of d f is x coz theta. so this can be written as d f x. coz theta.
if we substitute the value of d f. this will be d m omega square, x square sine theta.
coz theta. so if this is the torque acting, on the rod due to, the centrifugal force on
element about c. here we can write total torque, on rod. for all the centrifugal forces we
can write, as integration of d tau. and here d m also we can substitute as m by l d x.
then, this will give us. m by l omega square. x square, sine theta coz theta. d x. and we
integrated within limits from, minus l by 2 to, plus l by 2. here all constants can
be taken out, which are, m omega square by l. sine theta coz theta and this is integration
of x square. d x from minus l by 2 to, plus l by 2. which gives on integrating this becomes
x cube by 3. and this is m omega square by, l. sine theta coz theta multiplied by its
integration is x cube by 3 and, we substitute limits from. minus l by 2 to, plus l by 2.
if you numerically calculate this value after substitution of limits, then this will give
us 1 by 3, em omega square by l. sine theta coz theta multiplied by if we put the limits
then this will be l cube by 8. this minus minus plus, l cube by, 8. and, this will be
equals to. on solving we get 1 by 12. em omega square, l square, sine theta. coz theta which
is the result of this, problem.

1. Soumil Sahu said:

is it possible to solve this without integration?

December 18, 2016
2. sai charan reddy said:

ur wonderful I have never seen a professor like u physics galaxy will be world wide famous

January 3, 2017
3. Paras Rajput said:

sir in which video of your lecture I can find the instantaneous force by Hinge concept

January 13, 2017
4. prashant kumar said:
5. Evil Assassin said: