Does the block move along the pink dotted lines

[Kaushik]

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)

Is the figure above correct? If not, why?

 

[haruspex]

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.

 

[Kaushik]

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.

 

[haruspex]

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.

 

[Kaushik]

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?

 

[kuruman]

(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?

 

[Kaushik]

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?

 

[haruspex]

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.

 

[kuruman]

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."

 

[haruspex]

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?

 

[Kaushik]

Mechanical energy conservation

Using your hint, i found $V(Θ) = √(2gRcosΘ)$

Thanks a lot.

 

[kuruman]

Using your hint, i found $V(Θ) = √(2gRcosΘ)$

Thanks a lot.

Is that what you were looking for?

 

[Kaushik]

Is that what you were looking for?

Apart from that, I have 2 more questions to solve.

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.

ii) Normal reaction when the vertical component of the block's velocity is maximum.

This happens when the angle is 90 right ?

 

[kuruman]

This happens when the angle is 90 right ?

Look at your expression for $V(\Theta)$. At what angle does it have a maximum?

 

[Kaushik]

Look at your expression for $V(\Theta)$. At what angle does it have a maximum?

I meant 0.

 

[kuruman]

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.

 

[Kaushik]

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.

 

[kuruman]

That's it, good job. Now what do you think you should do next? What's the purpose of drawing a free body diagram?

 

[Kaushik]

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 $N = 3mg$?

 

[kuruman]

Yes and yes.
I stand corrected. I misread the question. See post #34 by @jbriggs444 who has a knack for keeping me honest.

 

[Kaushik]

Yes and yes.

Thanks a lot for your help.

 

[kuruman]

You are welcome. Can you finish the problem now and answer the third question?

 

[Kaushik]

You are welcome. Can you finish the problem now and answer the third question?

Yes.

 

[jbriggs444]

ii) Normal reaction when the vertical component of the block's velocity is maximum.

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.

 

[Kaushik]

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?

 

[jbriggs444]

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?

 

[Kaushik]

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?

 

[jbriggs444]

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.

 

[Kaushik]

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.

 

[kuruman]

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.

 

[Kaushik]

I'll try again, yes. But first you have to find the speed.

But we found speed to be $V_0 = √(2gRcosΘ)$ . Isn't?

 

[jbriggs444]

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: $\theta$

 

[Kaushik]

We have the speed: $\sqrt{2gR\cos \theta}$
We have the angle of the perpendicular: $\theta$

Is $V\sin{\theta}$ the vertical component?

 

[jbriggs444]

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.

 

[Kaushik]

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.

 

[kuruman]

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.

 

[Kaushik]

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 $\theta = \tan^{-1}(\sqrt2)$

 

[Kaushik]

So is this correct –> $N = mg + \frac{2mg}{\sqrt3}$

 

[kuruman]

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.

 

[kuruman]

So is this correct –> $N = mg + \frac{2mg}{\sqrt3}$

No. What is the equation for $N$ you get from the FBD at any angle $\theta$?

 

[Kaushik]

No. What is the equation for $N$ you get from the FBD at any angle $\theta$?

The centripetal acceleration is $\frac{2g}{\sqrt3}$, right?

 

[kuruman]

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.

 

[Kaushik]

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 $a_c=\frac{V_0^2}{R}$
$a_c = 2g\cos(\theta) = \frac{2g}{\sqrt3}$

 

[kuruman]

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.

 

[Kaushik]

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 $N = \sqrt{3} mg$.

 

[kuruman]

That is correct. How would you answer part (iii)? Another FBD is needed.

 

[Kaushik]

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.

 

[kuruman]

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?

 

[Kaushik]

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) ?

 

[kuruman]

$F_{net}$ (weight and normal) ?

Absolutely. Now draw the FBD.