Doubt by Nandkishor Rathod

Q.84 ???

Answers

Answer by

This is a very standard result which you should remember. we encounter this kind of heat loss in collision ( in 1 Dimension), , when two charged capacitors are joined together. All have similar expression.

(1/2)* I1I2/(I1 +I2) *(w1 - w2)^2  ( pls note that if w1 and w2 are in opposite direction, it will be (w1+w2)^2 )

 

We can prove this result by applying conservation of angular momentum. 

I1*w1 + I2*Iw2 = (I1 + I2)w

 

Then heat loss will be initial KE - Final KE =  (1/2)*I1*w1^2 + (1/2)*I2*w2^2 - (1/2)*(I1+I2)*(w1+w2)^2

Answer by

 (1/2)*I1*w1^2 + (1/2)*I2*w2^2 - (1/2)*(I1+I2)*(w1+w2)^2 

plugging the value of w from above equation I1*w1 + I2*Iw2 = (I1 + I2)w and solving, we get,

(1/2)* I1I2/(I1 +I2) *(w1 - w2)^2

Answer by Nandkishor Rathod

plugging the value of w from above equationΒ I1*w1 + I2*Iw2 = (I1 + I2)w and solving, we get, (1/2)* I1I2/(I1 +I2) *(w1 - w2)^2 I don't understand this step , please help me

Answer by

See, w is obtained from this equation I1*w1 + I2*Iw2 = (I1 + I2)w.

=> w = (I1*w1 + I2*Iw2 )/ (I1 + I2)

so, we can put the value of w in heat loss  =  (1/2)*I1*w1^2 + (1/2)*I2*w2^2 - (1/2)*(I1+I2)*(w1+w2)^2 

Then further simplifying, we will get. => heat loss = (1/2)* I1I2/(I1 +I2) *(w1 - w2)^2

Hope you understand it now. 

Answer by Nandkishor Rathod

yes , I understand !!! thank you again 😊😊😊

Answer by

this is great Nandkishor. Keep learning and practising questions πŸ˜ŠπŸ˜ŠπŸ˜Š.