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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

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

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4 Answers

VINOD KUMAR
14 Points
11 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.

Akash Kumar Dutta
98 Points
11 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.

Abhishekh kumar sharma
34 Points
10 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

Sandeep
38 Points
6 years ago
In part 1
total mass(M)=\int_{0}^{2L}mx               
M=2ml2
=>  m=\frac{M}{2L^{2}}
 
subsitute in I = 4ml4
 
=>   I=2Ml^{2} 
PART 2 (ALTERNATE METHOD)
 
we can find the COM of the given rod 
COM=\frac{\int_{0}^{2L}dm\ast x}{M}     dm= mx
 
=> COM=\frac{8mL^{3}}{3M}     ( m=\frac{M}{2L^{2}} )
 
COM=\frac{4L}{3}
 
from thoerem of parallel axes
 
I=ICOM + Mr2
=> ICOM=I- M\frac{4L}{3}^{2}
=> ICOM=\frac{2ML^{2}}{9}
 
=>  Iat mid point = ICOM + \frac{ML^{2}}{9}
 
  Iat mid point = \frac{ML^{2}}{3}

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