Does the block move along the pink dotted lines

[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.

 

[Kaushik]

Absolutely. Now draw the FBD.

I am getting $N\cos(\theta) = mg$, Am I correct? If yes, what should I do after this?

 

[kuruman]

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.

 

[Kaushik]

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?

 

[kuruman]

That is correct. Does the angle look familiar?

 

[Kaushik]

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

 

[kuruman]

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?

 

[Kaushik]

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.

 

[kuruman]

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.

Thanks a lot for your help @kuruman @jbriggs444

Thank you for an interesting post. I hadn't seen this variation before.