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
Menu
Pratik Somani Grade: 12
        

A rod of mass m length l and modulus Y is rotating about an axis passing through one of its end perpendicular to its length with angular velocity[Eqn]\omega[/Eqn]. Find the strain at the center of the rod.

7 years ago

Answers : (2)

Askiitians Expert Bharath-IITD
23 Points
										

Dear Pratik,


    We know that centrifugl force= mass * radius * angular velocity2


so we get F = m * l/2 * ω2


radius = l/2 beacuse the centre of mass of the rod is at l/2 distance from one the end.


we know that Y= longitudinal stress/longitudinal strain


long strain = long stress/ Y


    Δl/l           = F/ (A*Y)


        Where A is cross sectional area of the rod....


So there is some data still missing..


I hope this may help u to some extent...


I suggest u to post ur questions properly...


Please feel free to post as many doubts on our discussion forum as you can. If you find any question
Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We
are all IITians and here to help you in your IIT JEE preparation.


All the best  !!!




Regards,

Askiitians Experts

Adapa Bharath

7 years ago
Pratik Somani
16 Points
										

I forgot to mention it........the cross sectional area is A. But the 4 options are.......


1) (mω2L)/(4AY)


2) (3mω2L)/(4AY)


3) (mω2L)/(8AY)


4) (3mω2L)/(8AY)

7 years ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies
  • Complete Physics Course - Class 12
  • OFFERED PRICE: Rs. 2,756
  • View Details
Get extra Rs. 551 off
USE CODE: CART20
  • Complete Physics Course - Class 11
  • OFFERED PRICE: Rs. 2,968
  • View Details
Get extra Rs. 594 off
USE CODE: CART20

Ask Experts

Have any Question? Ask Experts

Post Question

 
 
Answer ‘n’ Earn
Attractive Gift
Vouchers
To Win!!! Click Here for details