Solving Pendulum Problem (b): Kinetic & Potential Energy

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The discussion revolves around solving a pendulum problem involving kinetic and potential energy. The initial calculation incorrectly assumes that the pendulum can complete a circle with zero speed at the top, leading to the wrong height calculation for the pivot point. The correct approach requires determining the minimum speed needed at the top of the swing, which involves applying Newton's second law and principles of circular motion. Additionally, there is a query about why kinetic energy at the highest point is not considered in the solution, prompting further exploration of energy conservation principles. Understanding these concepts is crucial for accurately solving the pendulum problem.
jack1234
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Hi, for this question
http://tinyurl.com/2qgco5
I have problem for question b

My attempts for question (b)
The potential energy of the pendulum when it is released from the horizontal position
=The kinetic energy of pendulum at lowest point
=mgL

In order to make it swing in a complete circle, it needs to reach the height 2(L-d)
hence mgL=mg(2(L-d)); and I get d=L/2

This is wrong, what it wants is d=3/5L, may I know what is my mistake, and what is the correct approach?
 
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jack1234 said:
In order to make it swing in a complete circle, it needs to reach the height 2(L-d)
hence mgL=mg(2(L-d)); and I get d=L/2
This assumes that the sphere can make a complete revolution about the peg with a speed of zero at the top. This is incorrect--figure out the minimum speed needed at the top. Hint: Consider Newton's 2nd law and circular motion.
 
Thanks!
 
See my response in the other thread.
 
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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