Does the block move along the pink dotted lines

In summary, the conversation is discussing a problem where a small block is projected from the bottom of a fixed sphere with a certain speed, causing it to move in a vertical circular path. The questions to be answered include finding the speed of the block as a function of the angle of deflection from the lowest vertical, the normal reaction when the vertical component of the block's velocity is maximum, and the angle between the thread and the lowest vertical when the total acceleration vector of the block is directed horizontally. The suggested approach involves using mechanical energy conservation and drawing a free body diagram for the block's path. The conversation also addresses issues with the initial diagram and provides a hint for solving the problem.
  • #36
Kaushik said:
I was trying to differentiate it and solving it for zero.

Ok I'll wait for @kuruman to reply.
I shall redeem myself with the help of Mathematica that verified the derivative.
$$\frac{d}{dx}(\sin x \sqrt{\cos x})=\cos^{3/2}x-\frac{\sin^2 x}{2\sqrt{\cos x}}$$
Let ##u=\cos x##. Then setting the derivative equal to zero gives
$$u^{3/2}-\frac{1-u^2 }{2u^{1/2}}=0~\rightarrow 2u^2-1+u^2=0$$
That's an easy one to solve.
 
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  • #37
kuruman said:
I shall redeem myself with the help of Mathematica that verified the derivative.
$$\frac{d}{dx}(\sin x \sqrt{\cos x})=\cos^{3/2}x-\frac{\sin^2 x}{2\sqrt{\cos x}}$$
Let ##u=\cos x##. Then setting the derivative equal to zero gives
$$u^{3/2}-\frac{1-u^2 }{2u^{1/2}}=0~\rightarrow 2u^2-1+u^2=0$$
That's an easy one to solve.
After differentiating the square of ## V_0 ## , I got ## \theta = \tan^{-1}(\sqrt2)##
 
  • #38
So is this correct –> ## N = mg + \frac{2mg}{\sqrt3}##
 
  • #39
That's another way to do it. You can verify that ##\arctan(\sqrt{2})=\arccos(1/\sqrt{3})##. I did it with calculator but it should be doable algebraically.
 
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  • #40
Kaushik said:
So is this correct –> ## N = mg + \frac{2mg}{\sqrt3}##
No. What is the equation for ##N## you get from the FBD at any angle ##\theta##?
 
  • #41
kuruman said:
No. What is the equation for ##N## you get from the FBD at any angle ##\theta##?
The centripetal acceleration is ## \frac{2g}{\sqrt3}##, right?
 
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  • #42
No. The centripetal acceleration is ##a_c=v^2/R##. Please write down the equation that you get from the FBD when you apply Newton's second law for any value of the angle ##\theta##. Once you have that, then you can substitute any specific ##\theta## value that you want.
 
  • #43
kuruman said:
No. The centripetal acceleration is ##a_c=v^2/R##. Please write down the equation that you get from the FBD when you apply Newton's second law for any value of the angle ##\theta##. Once you have that, then you can substitute any specific ##\theta## value that you want.
We know that, ## V_0 = \sqrt{2Rgcos(\theta)}##
As ##a_c=\frac{V_0^2}{R}##
##a_c = 2g\cos(\theta) = \frac{2g}{\sqrt3}##
 
  • #44
Kaushik said:
We know that, ## V_0 = \sqrt{2Rgcos(\theta)}##
As ##a_c=\frac{V_0^2}{R}##
##a_c = 2g\cos(\theta) = \frac{2g}{\sqrt3}##
That is all true for the specific angle in part (ii) of the problem. However, your answer in post #38, ##N=mg+\frac{2g}{\sqrt{3}}## tells me that you think that you can write ##N=mg+ma_c##. That is not correct and that is the reason I asked you to derive the general expression for ##N## from the FBD.
 
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  • #45
kuruman said:
That is all true for the specific angle in part (ii) of thr problem. However, your answer in post #38, ##N=mg+\frac{2g}{\sqrt{3}}## tells me that you think that you can write ##N=mg+ma_c##. That is not correct and that is the reason I asked you to derive the general expression for ##N## from the FBD.
I am really sorry. I considered weight along N to be Mg. It is ##Mg\cos(\theta)##. Now I am getting ## N = \sqrt{3} mg ##.
 
  • #46
That is correct. How would you answer part (iii)? Another FBD is needed.
 
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  • #47
Kaushik said:
iii)The angle between the thread and the lowest vertical at the moment when the total acc. vector of the block is directed horizontally.
In this question, the resultant of ## a_{tangential} ## and ## a_{centripetal} ## should be horizontal right.
 
  • #48
I prefer to think of it a bit differently to help with drawing the appropriate FBD. If the net acceleration is horizontal, what other vector quantity must be horizontal?
 
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  • #49
kuruman said:
I prefer to think of it a bit differently to help with drawing the appropriate FBD. If the net acceleration is horizontal, what other vector quantity must be horizontal?
## F_{net} ## (weight and normal) ?
 
  • #50
Kaushik said:
## F_{net} ## (weight and normal) ?
Absolutely. Now draw the FBD.
 
  • #51
kuruman said:
Absolutely. Now draw the FBD.
I am getting ## N\cos(\theta) = mg ##, Am I correct? If yes, what should I do after this?
 
  • #52
That is the condition, yes. How would you find the angle at which this is the case? Hint: I asked you to do something in post #42.
 
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  • #53
kuruman said:
That is the condition, yes. How would you find the angle at which this is the case? Hint: I asked you to do something in post #42.
## N\sin(\theta) ##'s component towards the centre is causing the centripetal acceleration.
So from the condition we got, ## N = mg/cos(\theta) ##.
Substituting the value of ## N ## in ## N\sin^2(\theta) = 2mg\cos(\theta) ##
We get, ##\theta = \tan^{-1}(\sqrt2) ##.
Is this correct?
 
  • #54
That is correct. Does the angle look familiar?
 
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  • #55
kuruman said:
That is correct. Does the angle look familiar?
Yes.
So the angle at which the velocity's vertical component is maximum is same as the angle at which the ## F_{net} ## is Horizontal.

Thanks a lot for your help @kuruman @jbriggs444
 
  • #56
Kaushik said:
Yes.
So the angle at which the velocity's vertical component is maximum is same as the angle at which the ## F_{net} ## is Horizontal.
In retrospect and with the benefit of 20/20 hindsight, do you see why it has to be this way?
 
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  • #57
kuruman said:
In retrospect and with the benefit of 20/20 hindsight, do you see why it has to be this way?
The velocity's vertical component will be maximum when acceleration along the vertical is minimum. So when ## a ## along vertical is 0, the net acceleration is horizontal.
 
  • #58
Kaushik said:
So when a a along vertical is 0, the net acceleration is horizontal.
This just says that when one component of the acceleration is zero, there is only the other component left.

Here is what I had in mind. The vertical component of the velocity has a maximum when the vertical component of acceleration is zero. When the vertical component of the acceleration is zero, the vertical component of ##F_{net}## is zero. The vertical component of ##F_{net}## is zero when the vertical component of ##N## is equal to the weight. Therefore, when the vertical component of ##N## is equal to the weight, the vertical component of the velocity is maximum and vice-versa.

Kaushik said:
Thanks a lot for your help @kuruman @jbriggs444
Thank you for an interesting post. I hadn't seen this variation before.
 
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