I have done in rotation a Problem where we put a rotating Disc on horizontal rough platform, find the Average Torque and then solve the equation.
Even the ring one is also easier.
But How do I do it when there is a square since the the limit of integrations will vary (Since Diagonal is longer) if we take the point of rotation at the center or at any corner?
62the inner circular part, we can integrate directly
The integral will come out to be slightly dirty.... based on theta... you can find the arc length...
If u get the hint try otherwise i will try to post the solution....
1corner should be taken ,as it is point of "Instantenous Rotation"
Can you please post the problem statement ?
71@kreyszig
You're misinterpreting what I've asked. Read my doubt atleast 10 times.
@Nishant Sir
"the inner circular part, we can integrate directly"
-> Yes, that is okay But that circular part would not cover ' that extra ' length along the diagonals. Or it would?
62@Vivek... I meant that inner part is easy to calculate...
The extra part is one where we have to integrate
71Q: A Square with dimension ' a ' is given an initial angular velocity ' w '. The square can rotate freely about the axis passing through one of its corners. It take time ' t ' to stop completely. Find the time taken to stop if the axis passes through the center of the square plate.
Take mass to be M.
1try to solve it by finding out the MOI about any one of the corners
The answer to your Q shud be = t/4 (if i hvn`t done my calculations wrong)
262i think it should be t/2 as moi about center is double of that from the corner..
1moi about the the centre is Ma2/6
now if u apply the parallel axis theorem then it`l come
= 2Ma2/3
@aditya....check ur answr
262i thought the axis is passing through both the corners.. my mistake.
1@ Debosmit: I dont believe that the ans should be t/4. As torque about centre and torque about corner wont be equal.
71Everyone is just calculating the MOI and not the total torque that would be acting.
Calculating MOI is not a problem here.
1if the MOI about the corner is 2/3ma2 now since the torque acting is constant we can apply,
w2=w1-\alphat
since w2 is 0.
hence,\alpha=w/t.
now,\Gamma=Ia
we knw I,we can find T.
