Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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

Grade:

4 Answers

VINOD KUMAR
14 Points
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.

Akash Kumar Dutta
98 Points
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.

Abhishekh kumar sharma
34 Points
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

Sandeep
38 Points
4 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}

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free