Tension of rope by hanging mass

AI Thread Summary
The discussion focuses on calculating the tension in a rope when an 11 kg mass is released from a quarter-circle position. Key forces acting on the mass include gravitational force and tension, with the mass accelerating downward due to gravity. At the bottom of its path, the tension in the rope must counteract both the weight of the mass and provide the necessary centripetal force for its circular motion. Utilizing force diagrams and energy equations is emphasized as essential for understanding the mechanics involved. The final tension can be calculated using these principles, highlighting the importance of applying physics concepts effectively.
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Homework Statement



A mass of 11 kg is hung on a rope of L = 2.7 meters. It is raised by 90 degrees (a quarter circle) held at rest, then released and it falls due to gravity alone. What is the tension in the rope at the bottom of its path in Newtons?
 
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What do you think?

Hint: What forces act on the mass? What is its acceleration?
 
force diagrams yay! i remember those helped me understand what's was going on.
 
Force Diagrams saves lives everyday. Don't forget to use energy equations for Mechanics they are also good for your health :)
 
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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