Solve Circular Motion Homework: Find Tension 1 & 2

AI Thread Summary
The discussion focuses on solving a circular motion homework problem involving a mass attached to two strings on a vertical shaft. The tangential velocity of the mass is given as 3.10 m/s, leading to an acceleration calculation of 2.48 m/s². The participant sets up equations for the forces in both the x and y directions, incorporating tensions T1 and T2 and the mass's weight. They express the equations for force balance and seek confirmation on their approach before solving for the tensions. The thread emphasizes the importance of correctly applying circular motion principles and resolving forces into components.
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Homework Statement



A shaft is set up in the vertical direction. Two strings are attached to a mass m, the shaft is rotates so that the tangential velocity of m is 3.10 m/s

a) Find tension of string 1 & 2

Homework Equations





The Attempt at a Solution



First I found the acceleration

a = v2/r = 3.12/1.25 = 2.48 m/s2

Fa = ma = 1.75*2.48 = 4.34 N

I then took it into components

\sumFx = Fa - T1sin\theta - T2sin\theta = 0 = 4.34 - T1sin30 - T2sin30


\sumFy = T1cos30 - mgT2cos30 = 0

are these correct i would just solve for the Ts here on after
 

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i thin Fy = T1 - T2 - mgcos30 = 0
 
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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