Minimum Acceleration for Ball to Reach Top of Vertical Ring

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The discussion centers on determining the minimum vertical acceleration required for a ball to reach the top of a vertical ring with radius R. The initial attempt used energy conservation principles, leading to an incorrect acceleration value of 3g/4, while the textbook states the correct answer is 4g/5. Participants emphasized the importance of considering centripetal acceleration and the implications of the ball's speed at the top of the ring. Acknowledgment of the need for a sufficient energy state to prevent the ball from falling back before reaching the top was also highlighted. The conversation concludes with a recognition that centripetal acceleration is crucial for solving the problem correctly.
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Homework Statement


Small ball moves on the inner surface of the vertical ring with radius R. Moving ball reaches maximum height equal to R/2. What minimum acceleration (in vertical direction) is required (to the system of ring and ball) to make the ball reach the top of the ring?


Homework Equations


All are provided in my solution... there might be another solving methods I haven't tried


The Attempt at a Solution


I've tried to apply energy conservation law in this situation: ball has kinetic energy Ek=m(v^2)/2 in the bottom of the ring. All kinetic energy is converted to potential energy when ball reaches the top position (height equal to R/2). I wrote down energy conservation law: m(v^2)/2=mg(R/2) ---> v^2=gR; When we give vertical acceleration a to the system, acting force is equal to m(g-a), not mg. Value of a must satisfy the condition that ball reaches top of the ring (height 2R). Then I've written down again: m(v^2)/2=2m(g-a)R ---> v^2=4(g-a)R ---> gR=4(g-a)R---> a = 3g/4; However, correct answer provided in my textbook is a=4g/5. I can't understand what's wrong with my solution... I would be very thankful for your help!
 
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Welcome to PF!

Hi PipelineDream! Welcome to PF! :smile:

It's just a sneakier version of those rollercoaster problems …

you haven't taken into account the fact that if the ball only has enough energy to approach the top at zero speed, it will have fallen into the middle long before it gets there! :rolleyes:

Use centripetal acceleration ! :wink:
 


tiny-tim said:
Hi PipelineDream! Welcome to PF! :smile:

It's just a sneakier version of those rollercoaster problems …

Not exactly. I mean, you don't typically take an entire roller coaster and its track and put it in a gigantic elevator that accelerates upwards at acceleration "a," do you? This is what the problem is saying. If there is enough energy in the system for the ball to oscillate back and forth in the loop up to height R/2 on each side when the whole system is stationary in the Earth's reference frame, then what happens if you put the track + ball in a reference frame that is accelerating upwards? I have to admit that right now I'm not sure.


tiny-tim said:
you haven't taken into account the fact that if the ball only has enough energy to approach the top at zero speed, it will have fallen into the middle long before it gets there! :rolleyes:

I'm dubious. Can you explain why this is true and what is wrong with the conservation of energy argument in that instance?
 
Hi cepheid! :smile:
cepheid said:
I'm dubious. Can you explain why this is true and what is wrong with the conservation of energy argument in that instance?

'cos …

i] the reaction force will be zero well before the top, and

ii] it gives the right answer! :biggrin:
 
Hey, I see centripetal acceleration works very well :wink: Thanks very much for your advises, they really helped me!
 
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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