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# a non uniform rod AB has a mass M and length 2L the mas per unit length of the rod is mx at a point of the rod distant x from A.find the moment of inertia of this rod about an axis perpendicular to the rod (!)through A (2)through the mid point of AB

8 years ago

consider a strip of the rod ''dx'' at a length ''x'' from A.mass of the small strip is mx*dx.moment of inertia of the small strip about A IS m*x*dx*x*x.on integrating we get moment of inetia about A as 4*m*L*L*L*L.

ABOUT MID POINT OF AB.consider a strip of length ''dx'' at a distance of x from mid point P.mass of it is m*(L-x)*dx.moment of inertia of half of the rod about P is m*(L-x)*dx*x*x.on integration we get moment of inertia of half of the rod.similarly do for the other half.adding them we get the answer.

8 years ago

Dear Park kl,

for part 1.

mass of part dx at a distance x from A=mxdx
2L
I= integrating... [ mx.x^2.dx]= 4mL^4
0

for part 2.

let the midpoint be C.then.let the part whose mass per unit length is mentioned be at a distance of x from C.then

distance from A=L-x
L
I= integrating... [m(L-x).x^2dx]= m(L^4)/6.
-L

(ANS).

Regards,
Akash.

8 years ago

for part 1.

mass of part dx at a distance x from A=mxdx
2L
I= integrating... [ mx.x^2.dx]= 4mL^4
0

for part 2.

let the midpoint be C.then.let the part whose mass per unit length is mentioned be at a distance of x from C.then

distance from A=L-x
L
I= integrating... [m(L-x).x^2dx]= m(L^4)/6.
-L

4 years ago
In part 1
total mass(M)=mx
M=2ml2
=>  m=

subsitute in I = 4ml4

=>
PART 2 (ALTERNATE METHOD)

we can find the COM of the given rod
COM=     dm= mx

=> COM=     ( m= )

COM=

from thoerem of parallel axes

I=ICOM Mr2
=> ICOM=I- M
=> ICOM=

=>  Iat mid point = ICOM +

Iat mid point =