Question

Please someone help to solve the sum of question no. 2

Answer 1

Answer:

Hi !

According to the question as one fourth of the whole slab mass is being removed finally , the mass of that one fourth part will surely be one fourth of the total mass.

Now, I'll denote the centre of mass of individual quadrants of the given mass as

(x,y) : < x1, y1 > rather than using different equations for both x and y dimensions separately.

We know,

r(com) = € miri/€mi

or,

x(com) = €mixi/€mi

and similarly for other dimensions.

Now,

in this case as that one fourth mass has been removed we can take it as a negative mass (-m/4) if we suppose the initial mass of the whole slab to be considered as m.

So,

r(com) = m <a/2,b/2> - m/4 <a/2+a/4,b/2+b/4> / m-m/4

= m<a/2,b/2> - m/4<3a/4,3b/4> / 3m/4

= < ma/2 - 3ma/16 , mb/2 - 3mb/16 > / 3m/4

= < 5a/12 , 5b/12 >

Hope this helps you ! ｡◕‿◕｡