Work and gravitational potental energy

AI Thread Summary
The discussion focuses on a physics problem involving a thin rod and a ball, exploring the work done by gravitational force and the change in gravitational potential energy as the ball descends. The calculation for work done is derived using the formula W = mgh, where h is the height difference determined by the angle of descent. The term 1 - cos(30) represents the vertical displacement of the ball relative to its initial position. Participants clarify that the work done is calculated by comparing the gravitational potential energy at two points in the motion. The thread emphasizes understanding the relationship between work, energy, and the geometry of the motion involved.
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Homework Statement


A thin rod, of length L = 2.00 m and negligible mass, that can pivot about one end to rotate in a vertical circle. A ball of mass m = 5.00 kg is attached to the other end. The rod is pulled aside to angle theta = 30 and released with initial velocity v= 0. As the ball descends to its lowest point, (a) how much work does the gravitational force do on it and b) what is the change in the gravitational potental energy of the ball - Earth system?.. etc.


Homework Equations





The Attempt at a Solution


In the solution, they show W = 5 x 9.8 x 2(1- cos30)

where do they get the 1-cos30??
 
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W = Work at one step- Work at another step

= 5 x 9.8 x 2(1) - 5 x 9.8 x 2(cos30)
 
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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