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.
  • #1
Kaushik
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17
Homework Statement
A small block ##m## is projected with speed ##v_0 = \sqrt{2gR}## from bottom of a fixed sphere of radius ##R## , so that the block moves in a vertical circular path.
Relevant Equations
Attached below.
Does the block move along the pink dotted lines as attached in the figure below?
I tried to draw the FBD of the small block ##m ## at the lowermost point which is also attached below.(The direction of ## v_0 ## is actually tangential)
phy.forum.5.jpg


Is the figure above correct? If not, why?
 
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  • #2
Kaushik said:
Problem Statement: A small block ## m ## is projected with speed ## v_0 = √2gR ## from bottom of a fixed sphere of radius ##R## , so that the block moves in a vertical circular path.
Relevant Equations: Attached below.

Does the block move along the pink dotted lines as attached in the figure below?
I tried to draw the FBD of the small block ##m ## at the lowermost point which is also attached below.(The direction of ## v_0 ## is actually tangential)
View attachment 246940

Is the figure above correct? If not, why?
Your diagram is fine for the initial motion, but the speed will drop as the block rises. What exactly is the question to be answered? You probably need an FBD for a more general position.
 
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  • #3
haruspex said:
What exactly is the question to be answered?

The questions to be answered are -
i) The speed of the block as a function of ## Θ ##, where Θ is the angle of deflection from the lowest vertical.
ii) Normal reaction when the vertical component of the block's velocity is maximum.
iii)The angle between the thread and the lowest vertical at the moment when the total acc. vector of the block is directed horizontally.


I didn't write these questions before as I wanted the image.Like, I wanted the intuition of how this system really looks.

Thanks.
 
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  • #4
Kaushik said:
The questions to be answered are -
i) The speed of the block as a function of ## Θ ##, where Θ is the angle of deflection from the lowest vertical.
ii) Normal reaction when the vertical component of the block's velocity is maximum.
iii)The angle between the thread and the lowest vertical at the moment when the total acc. vector of the block is directed horizontally.


I didn't write these questions before as I wanted the image.Like, I wanted the intuition of how this system really looks.

Thanks.
Then, as I posted, you need to draw the FBD and write equations for the general position, as described in (i).
Don't bother trying to draw perspective, as you seem to have done in post 1. Just show the plane in which the block moves.
 
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  • #5
Kaushik said:
deflection from the lowest vertical.
What does the above line mean? Should I assume that the block lies at an angle theta from the lowest point?
 
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  • #6
Kaushik said:
(The direction of ## v_0 ## is actually tangential)
Is the figure above correct? If not, why?
The diagram does not look correct. Since the initial velocity is tangent to the circle, the block will start sliding on the inside surface of the circle, presumably without friction. The questions to consider are, will the block slide all the way up to the "12 o'clock" position or will it fall off at some intermediate point and, if so, where is that point and what is the subsequent path of the block?
 
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  • #7
kuruman said:
The diagram does not look correct. Since the initial velocity is tangent to the circle, the block will start sliding on the inside surface of the circle, presumably without friction. The questions to consider are, will the block slide all the way up to the "12 o'clock" position or will it fall off at some intermediate point and, if so, where is that point and what is the subsequent path of the block?
We know that the velocity decreases as the angle theta increases because of ## g ##. Can we find the velocity of the block as a function of theta? If yes, can I get a hint?
 
  • #8
Kaushik said:
What does the above line mean? Should I assume that the block lies at an angle theta from the lowest point?
The "lowest vertical " means the radius line from the centre of the circle to the bottom of the circle. Take another radius line from the centre of the circle to the block and call the angle between the two radii theta,
Draw the FBD for that.
 
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  • #9
Hint: Mechanical energy conservation plus a free body diagram. I assume the question to be answered is "Draw the complete path of the block inside the shell."
 
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  • #10
Kaushik said:
We know that the velocity decreases as the angle theta increases because of ## g ##. Can we find the velocity of the block as a function of theta? If yes, can I get a hint?
Anything conserved?
 
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  • #11
kuruman said:
Mechanical energy conservation
Using your hint, i found ##V(Θ) = √(2gRcosΘ)##

Thanks a lot.
 
  • #12
Kaushik said:
Using your hint, i found ##V(Θ) = √(2gRcosΘ)##

Thanks a lot.
Is that what you were looking for?
 
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  • #13
kuruman said:
Is that what you were looking for?
Apart from that, I have 2 more questions to solve.
Kaushik said:
ii) Normal reaction when the vertical component of the block's velocity is maximum.
iii)The angle between the thread and the lowest vertical at the moment when the total acc. vector of the block is directed horizontally.
These are the remaining questions I need to solve.
Kaushik said:
ii) Normal reaction when the vertical component of the block's velocity is maximum.
This happens when the angle is 90 right ?
 
  • #14
Kaushik said:
This happens when the angle is 90 right ?
Look at your expression for ##V(\Theta)##. At what angle does it have a maximum?
 
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  • #15
kuruman said:
Look at your expression for ##V(\Theta)##. At what angle does it have a maximum?
I meant 0.
 
  • #16
Kaushik said:
I meant 0.
Good. Now draw a free body diagram of the block wjen it is at ##\Theta=0##. However, before you do that count how many items outside the block exert a force on it. You should draw as many forces acting on the block as you have items interacting with it. Also, draw the acceleration of the block outside the FBD.
 
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  • #17
kuruman said:
Good. Now draw a free body diagram of the block when it is at ##\Theta=0##. However, before you do that count how many items outside the block exert a force on it. You should draw as many forces acting on the block as you have items interacting with it. Also, draw the acceleration of the block outside the FBD.
The forces I noticed are - Normal Force( Centripetal force here), Weight.
 
  • #18
That's it, good job. Now what do you think you should do next? What's the purpose of drawing a free body diagram?
 
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  • #19
kuruman said:
That's it, good job. Now what do you think you should do next? What's the purpose of drawing a free body diagram?
Now should I equate ## F_{net} ## to centripetal acceleration? Is it ## N = 3mg ##?
 
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  • #20
Yes and yes.
I stand corrected. I misread the question. See post #34 by @jbriggs444 who has a knack for keeping me honest.
 
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  • #21
kuruman said:
Yes and yes.
Thanks a lot for your help.
 
  • #22
You are welcome. Can you finish the problem now and answer the third question?
 
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  • #23
kuruman said:
You are welcome. Can you finish the problem now and answer the third question?
Yes.
 
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  • #24
Kaushik said:
ii) Normal reaction when the vertical component of the block's velocity is maximum.
Kaushik said:
Now should I equate ##F_{net}## to centripetal acceleration? Is it ##N=3mg##?
No and no.

##F_{net}## is only equal to [mass times] centripetal acceleration if tangential acceleration is zero. The vertical component of the block's velocity will not be maximized at the bottom of the circle.
 
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  • #25
jbriggs444 said:
No and no.

##F_{net}## is only equal to [mass times] centripetal acceleration if tangential acceleration is zero. The vertical component of the block's velocity will not be maximized at the bottom of the circle.
If not at the bottom, where else?
 
  • #26
Kaushik said:
If not at the bottom, where else?
You have speed as a function of angle. What is the vertical component of velocity as a function of angle?
 
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  • #27
jbriggs444 said:
You have speed as a function of angle. What is the vertical component of velocity as a function of angle?
Can I get a hint?
 
  • #28
Kaushik said:
Can I get a hint?
Umm, it is pretty simple. If you know how fast something is moving and in what direction it is moving, surely you can figure out the vertical component of that motion.
 
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  • #29
jbriggs444 said:
Umm, it is pretty simple. If you know how fast something is moving and in what direction it is moving, surely you can figure out the vertical component of that motion.
So the direction is tangential. So I have to find its vertical component.
 
  • #30
Kaushik said:
So the direction is tangential. So I have to find its vertical component.
I'll try again, yes. But first you have to find the speed.
 
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  • #31
kuruman said:
I'll try again, yes. But first you have to find the speed.
But we found speed to be ## V_0 = √(2gRcosΘ) ## . Isn't?
 
  • #32
kuruman said:
I'll try again, yes. But first you have to find the speed.
We have the speed: ##\sqrt{2gR\cos \theta}##
We have the angle of the perpendicular: ##\theta##
 
  • #33
jbriggs444 said:
We have the speed: ##\sqrt{2gR\cos \theta}##
We have the angle of the perpendicular: ##\theta##
Is ## V\sin{\theta}## the vertical component?
 
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  • #34
Kaushik said:
Is ## V\sin{\theta}## the vertical component?
Yup. So now you have the hard part. Finding a maximum for an ugly looking formula. I do not see an easy approach short of trial and error. Finding the maximum by taking the derivative and solving for a zero does not look good. Maybe @kuruman has some algebra trick up his sleeve.
 
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  • #35
jbriggs444 said:
Yup. So now you have the hard part. Finding a maximum for an ugly looking formula. I do not see an easy approach short of trial and error. Finding the maximum by taking the derivative and solving for a zero does not look good. Maybe @kuruman has some algebra trick up his sleeve.
I was trying to differentiate it and solving it for zero.

Ok I'll wait for @kuruman to reply.
 
<h2>1. What is the purpose of the pink dotted lines?</h2><p>The pink dotted lines represent the intended path for the block to move along. They serve as a visual guide for the movement of the block.</p><h2>2. How does the block move along the pink dotted lines?</h2><p>The block moves along the pink dotted lines by following a set of instructions or commands, such as pushing or pulling it along the path.</p><h2>3. Can the block move in any direction along the pink dotted lines?</h2><p>It depends on the specific instructions given to the block. Some blocks may only be able to move in a straight line, while others may have the ability to turn or change direction along the pink dotted lines.</p><h2>4. What happens if the block deviates from the pink dotted lines?</h2><p>If the block deviates from the pink dotted lines, it may not reach its intended destination or complete its designated task. It is important for the block to stay on the pink dotted lines for efficient and accurate movement.</p><h2>5. Are the pink dotted lines necessary for the block to move?</h2><p>No, the pink dotted lines are not necessary for the block to move. They are simply a visual aid to help guide the movement of the block along a specific path.</p>

1. What is the purpose of the pink dotted lines?

The pink dotted lines represent the intended path for the block to move along. They serve as a visual guide for the movement of the block.

2. How does the block move along the pink dotted lines?

The block moves along the pink dotted lines by following a set of instructions or commands, such as pushing or pulling it along the path.

3. Can the block move in any direction along the pink dotted lines?

It depends on the specific instructions given to the block. Some blocks may only be able to move in a straight line, while others may have the ability to turn or change direction along the pink dotted lines.

4. What happens if the block deviates from the pink dotted lines?

If the block deviates from the pink dotted lines, it may not reach its intended destination or complete its designated task. It is important for the block to stay on the pink dotted lines for efficient and accurate movement.

5. Are the pink dotted lines necessary for the block to move?

No, the pink dotted lines are not necessary for the block to move. They are simply a visual aid to help guide the movement of the block along a specific path.

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