Calculating Swinger Mass with Limited Information

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To calculate the maximum mass of the swinger given a rope tension of 800N and an angle of 27 degrees from vertical, the tension must balance both the gravitational force and the centripetal force. The equation T = mg + mv²/r can be used, but without specific values for height, velocity, or radius, direct calculations are challenging. By labeling the rope length as "r" and the mass as "m," these variables can be eliminated in the final calculations. This approach allows for solving the problem despite limited information. Understanding the relationship between tension, mass, and angles is crucial in this scenario.
1MileCrash
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A rope has a max tension of 800N.

A swinger is pulled back so that the rope is 27 degrees from vertical.

What's the max mass of the swinger?

Ok... I have tried a few approaches. I can't relate potential to kinetic because I am not given height. And even if i could get to kinetic, i don't know mass!

I figure that tension must be able to support weight and apply the centripetal force,

T=mg + mv^2/r

But I have no r.. no length, no v.. what's the key?
 
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Hi 1MileCrash! :smile:

Call the length "r" and the mass "m" …

they'll cancel out in the end! :wink:​
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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