Question

\[T=\int\limits_{A}^{}rdF=\int\limits_{A}^{}2\tau 2 \pi r dr\] Can someone explain how we switched to dr please. that tau is = to shear stress

Answer 1

τ=Force/Area

Answer 2

please doctor im damaged

Answer 3

write the complete question please

Answer 4

looks like in your question, the area on which the force is applied changes, that is why you have dr

Answer 5

\[\tau= F/A\]
\[F=\tau A\]
derivative F in r

\[dF/dr = A \tau\]
\[dF = A \tau dr\]

then insert to \[\int\limits_{A}^{} A \tau r dr\]

and then you solve it by ur self