Click to Chat
1800-2000-838
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
A square of side 4 units is divided into four equal squares. the square portion ABCH is removed and placed over HCDE. What is the new centre of mass of the system
Dear Sai, First we can consider the square slab is placed such that its centre of mass is at C(0,0) Now we can consider this centre of mass to be contributed by the square slab ABCH and the remaining part HCBGIFDE and let the position of centre of mass of this part is (x1, y1) and the removed square part will be (1,1) From these we can find the position of th Centre of mass of the part remaining after removing the square ABCH xcm = ( m 1 x 1+ m 2 x 2 )/m 1 + m 2 ycm = ( m 1 y 1 + m 2 y2 )/ m 1+ m 2 Where m1 = mass of HCBGIFDE = (Area of it) * thicknesss * density of the slab = A1 * t * d where A1 = 3* area of each square = 12 And similarly m2 = 4 * t * d And ( x2, y2 ) = ( 1, 1 ) and (xcm ,ycm) = ( 0,0 ) thus we would get ( x1, y1 ) = ( -1/3, -1/3 ) Now if the removed plate is kept on the Plate HCDE then its centre of mass becomes ( x2, y2 ) = ( -1, 1 ) Now we again calculate the centre of mass of the system bye using the formula xcm = ( m 1 x 1+ m 2 x 2 )/m 1 + m 2 ycm = ( m 1 y 1 + m 2 y2 )/ m 1 + m 2 we get the centre of mass of the system to be ( -1/2, 0 ) 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
Dear Sai,
First we can consider the square slab is placed such that its centre of mass is at C(0,0)
Now we can consider this centre of mass to be contributed by the square slab ABCH and the remaining part HCBGIFDE
and let the position of centre of mass of this part is (x1, y1) and the removed square part will be (1,1)
From these we can find the position of th Centre of mass of the part remaining after removing the square ABCH
xcm = ( m 1 x 1+ m 2 x 2 )/m 1 + m 2
ycm = ( m 1 y 1 + m 2 y2 )/ m 1+ m 2
Where m1 = mass of HCBGIFDE = (Area of it) * thicknesss * density of the slab = A1 * t * d
where A1 = 3* area of each square = 12
And similarly m2 = 4 * t * d
And ( x2, y2 ) = ( 1, 1 )
and (xcm ,ycm) = ( 0,0 )
thus we would get ( x1, y1 ) = ( -1/3, -1/3 )
Now if the removed plate is kept on the Plate HCDE
then its centre of mass becomes ( x2, y2 ) = ( -1, 1 )
Now we again calculate the centre of mass of the system bye using the formula
ycm = ( m 1 y 1 + m 2 y2 )/ m 1 + m 2
we get the centre of mass of the system to be ( -1/2, 0 )
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
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -