Optimizing Power Transmission with Differentiation: Solving for Maximum Velocity

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SUMMARY

The discussion focuses on optimizing power transmission in a belt system using differentiation. The power transmitted, represented by the equation P(v) = Tv - mv³, reaches its maximum when the velocity v is equal to √(T/3m). The differentiation process reveals that the maximum occurs when the derivative P' equals zero, leading to the equation 0 = T - 3mv². The solution confirms that the correct expression for maximum velocity is v = √(T/3m).

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Homework Statement


(a) A belt of mass m per unit length is wound partly around a pulley. The power P(v) transmitted is given by P=Tv-mv3 where T is the tension and v is velocity of the belt. Use differentiation to show that the maximum transmission occurs when v=sqrt(T/3m).


Homework Equations





The Attempt at a Solution


I differentiated to get P'=T-3mv2
then max occurs when P'=0
so 0=T-3mv2, run into trouble here:
v_max=+-sqrt(-T/3m)
how do I get rid of that minus T?

thanks
 
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You're able to differentiate, then you have problems using baisc algebra... I'm puzzled.

0=T-3mv^2

Move 3mv^2 on the other side, change sign !

3mv^2=T

v^2=\frac{T}{3m}

v=\sqrt\frac{T}{3m}
 
s**t, now I look stupid!
Thanks - guess I had a bit of a brain explosion.
 

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