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.

is it possible to solve this without integration?

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

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

R/Sir, can we solve the other question of same type if the axis is not vertical means for slanted axis by this method

Is this the same question as in ie irodov and also which came in aiims exam