Hard kinematics problem: Block pushed onto a moving conveyor belt

[IamVector]

so now how do we convert into lab frame?? by using hint but what is the logic behind it??

 

[PeroK]

so now how do we convert into lab frame?? by using hint but what is the logic behind it??

I've done too much for you already. The rules here are that we should only give hints.

You can work out what $\theta'$ is if you do your own diagram.

Whenever there are two frames, you need to think carefully about notation.

The hint was a general tip when you want to minimise a speed. Minimising $v^2$ avoids having to differentiate terms involving square roots. It's just a bit neater and quicker.

 

[IamVector]

Ooops. That's a double negative there in the middle. You screwed up the signs again.

That's a good argument against folding the sign on a component into the plus sign between the components. Do all of your signs in the component values and you won't get so confused.

sry
(-1+μgt/√5 )i+(2-2μgt/√5)j??

 

[jbriggs444]

The block starts at some initial speed $u'$ at some intial angle $\theta'$. The speed at time $t$ is given by $v' = u' - at$, where $a$ is the deceleration due to friction. $a = \mu g$

The angle remains constant, so we have:
$$v'_x = -v'\sin \theta', \ \ v'_y = v'\cos \theta'$$

Let's read through what @PeroK has written and try to decode the sign convention.

We know that we are using the belt frame from several hints. First, the block arrives at an angle. If we were using the ground frame, it would arrive purely crosswise. Second, we are using $v'$ to denote the velocity. One would expect to use unprimed velocities for the frame we really care about and primed velocities for the frame we are temporarily working in. Finally, we are told that the "angle remains constant". That only happens in the belt frame. In the ground frame, the object is going to be curving.

I am betting that the x-axis is laid out on the axis of the conveyer belt and the y-axis is laid out across the conveyer belt. We are notionally standing on the right side of the conveyer belt watching it move to the right in front of us. The positive x-axis extends to our right in the direction the belt is moving. The positive y-axis extends across the belt in front of us.

However, since we have adopted a frame of reference in which the belt is motionless, we are using primed coordinates. We have notionally stepped up onto the belt and are moving along with it.

An object enters the belt, passing between our two feet. It makes an angle $\theta'$ counter-clockwise from the y axis.

Now we check this interpretation. The x velocity of the object is $-v' \sin \theta'$. Yes, that works out. we are measuring the angle theta counter-clockwise.

The y velocity of the object is $v' \cos \theta'$. Yes, that works out the same way.

Now then you had asked:

initial angle between belt velocity and block velocity??

As @PeroK has hinted, the angle that a trajectory makes will depend on who is doing the measuring.

If you were riding in a railroad car throwing a ball up and then catching it again, you would say that the ball is being thrown at an angle of 90 degrees relative to the car's motion -- purely vertical. But a passerby would say that the ball is thrown at some angle forward of the vertical.

That is why @PeroK was careful to label the angle $\theta'$ rather than $\theta$. It is an angle measured from the belt frame.

As above, it appears to be the angle that the object's trajectory makes (in the belt frame) compared to a reference angle going straight across the belt.

 

[IamVector]

Let's read through what @PeroK has written and try to decode the sign convention.

We know that we are using the belt frame from several hints. First, the block arrives at an angle. If we were using the ground frame, it would arrive purely crosswise. Second, we are using $v'$ to denote the velocity. One would expect to use unprimed velocities for the frame we really care about and primed velocities for the frame we are temporarily working in. Finally, we are told that the "angle remains constant". That only happens in the belt frame. In the ground frame, the object is going to be curving.

I am betting that the x-axis is laid out on the axis of the conveyer belt and the y-axis is laid out across the conveyer belt. We are notionally standing on the right side of the conveyer belt watching it move to the right in front of us. The positive x-axis extends to our right in the direction the belt is moving. The positive y-axis extends across the belt in front of us.

However, since we have adopted a frame of reference in which the belt is motionless, we are using primed coordinates. We have notionally stepped up onto the belt and are moving along with it.

An object enters the belt, passing between our two feet. It makes an angle $\theta'$ counter-clockwise from the y axis.

Now we check this interpretation. The x velocity of the object is $-v' \sin \theta'$. Yes, that works out. we are measuring the angle theta counter-clockwise.

The y velocity of the object is $v' \cos \theta'$. Yes, that works out the same way.

Now then you had asked:

As @PeroK has hinted, the angle that a trajectory makes will depend on who is doing the measuring.

If you were riding in a railroad car throwing a ball up and then catching it again, you would say that the ball is being thrown at an angle of 90 degrees relative to the car's motion -- purely vertical. But a passerby would say that the ball is thrown at some angle forward of the vertical.

That is why @PeroK was careful to label the angle $\theta'$ rather than $\theta$. It is an angle measured from the belt frame.

As above, it appears to be the angle that the object's trajectory makes (in the belt frame) compared to a reference angle going straight across the belt.

I think I need to learn more before solving these problems thank you for your efforts btw it will be a great help if you could kindly tell me your approach of doing maybe it will work out for me or you could explain me the hint provided and logic behind it I am really confused with that .

 

[jbriggs444]

I think I need to learn more before solving these problems thank you for your efforts btw it will be a great help if you could kindly tell me your approach of doing maybe it will work out for me or you could explain me the hint provided and logic behind it I am really confused with that .

You are going to need to show some effort or put some better focus into your questions. I am not going to simply shotgun hints until one sticks.

 

[IamVector]

You are going to need to show some effort or put some better focus into your questions. I am not going to simply shotgun hints until one sticks.

I think my concepts are a bit weak this ques is from jaan kalda kinematics pdf I tried doing it to make my concepts strong to reach level of Olympiads question but I managed to do only first 5 rest are out of my range thanks for your help.

 

[IamVector]

You are going to need to show some effort or put some better focus into your questions. I am not going to simply shotgun hints until one sticks.

btw any advice from your side on improving my conceptual knowledge will be a great help

 

[PeroK]

I think my concepts are a bit weak this ques is from jaan kalda kinematics pdf I tried doing it to make my concepts strong to reach level of Olympiads question but I managed to do only first 5 rest are out of my range thanks for your help.

This is not an easy question. There are two things. One is notational technique. Look at the way that @jbriggs444 understood my notation and why I had used the variables I did. That's something to work on.

The other thing is not needing a one-step solution. Easy mechanics questions are a bit like doing chess puzzles where you need to find checkmate in 1-2 moves. A problem like this is more like finding a checkmate in 4-5 moves. You need to keep going with a line of thought, analysis and mathematics and see where it leads. And, judge when it's not leading anywhere and backtrack.

Both of these things require practice and a certain discipline to stick with an idea and think it through.

 

[IamVector]

This is not an easy question. There are two things. One is notational technique. Look at the way that @jbriggs444 understood my notation and why I had used the variables I did. That's something to work on.

The other thing is not needing a one-step solution. Easy mechanics questions are a bit like doing chess puzzles where you need to find checkmate in 1-2 moves. A problem like this is more like finding a checkmate in 4-5 moves. You need to keep going with a line of thought, analysis and mathematics and see where it leads. And, judge when it's not leading anywhere and backtrack.

Both of these things require practice and a certain discipline to stick with an idea and think it through.

thanks I will try it one more time

 

[PeroK]

thanks I will try it one more time

A third point is being able to work algebraically. You could certainly plug the numbers in in this case. But, I find the more practice you get with the algebraic patterns the better.

My solution is just over a page of A4. About 15-20 lines. So, you'll have to stick with it. Unless you do something very clever, the answer won't come out in a few lines.

That said, I suspect there may be a very neat, quick solution to this. But, I don't see it.

 

[jbriggs444]

A third point is being able to work algebraically. You could certainly plug the numbers in in this case. But, I find the more practice you get with the algebraic patterns the better.

My solution is just over a page of A4. About 15-20 lines. So, you'll have to stick with it. Unless you do something very clever, the answer won't come out in a few lines.

I may be streamlining out some steps with my intuition leading me helter skelter to a nice diagram. But the geometric approach allows a result to be read off almost immediately.

 

[IamVector]

A third point is being able to work algebraically. You could certainly plug the numbers in in this case. But, I find the more practice you get with the algebraic patterns the better.

My solution is just over a page of A4. About 15-20 lines. So, you'll have to stick with it. Unless you do something very clever, the answer won't come out in a few lines.

That said, I suspect there may be a very neat, quick solution to this. But, I don't see it.

ok thinking on it

 

[IamVector]

I may be streamlining out some steps with my intuition leading me helter skelter to a nice diagram. But the geometric approach allows a result to be read off almost immediately.

what is your approach??

 

[jbriggs444]

what is your approach??

Look at post #20 in this thread.

Following the approach means being able to understand that shifting from the belt frame to the ground frame amounts to adding a fixed rightward velocity (1 m/s in this case) to everything.

 

[IamVector]

Look at post #20 in this thread.

Following the approach means being able to understand that shifting from the belt frame to the ground frame amounts to adding a fixed rightward velocity (1 m/s in this case) to everything.

as you said to convert into lab frame I have to add 1m/s to everything I added it to my eq which will be equal to my velocity from lab frame(i think)

 

[jbriggs444]

as you said to convert into lab frame I have to add 1m/s to everything I added it to my eq which will be equal to my velocity from lab frame(i think)

Can we see what you have done so that we can critique and possibly guide?

 

[PeroK]

It's become a contest bewteen inspiration and geometric insight on the one hand and honest algebraic toil on the other!

 

[IamVector]

It's become a contest bewteen inspiration and geometric insight on the one hand and honest algebraic toil on the other!

I think I got the answer! (μgt/√5 )i+(2-2μgt/√5)j got this equation after adding 1 then took its mod thus forming a quadratic in μgt now to find the min value of vector we need min value of quadratic so I find its vertex and took out its min value ie (4/5)^1/2 = (2)/(5^1/2)

 

[PeroK]

I think I got the answer! (μgt/√5 )i+(2-2μgt/√5)j got this equation after adding 1 then took its mod thus forming a quadratic in μgt now to find the min value of vector we need min value of quadratic so I find its vertex and took out its min value ie (4/5)^1/2 = (2)/(5^1/2)

$\frac 2 {\sqrt 5} m/s$ is the answer I get.

In general the minimum speed is $v_b \cos \theta'$, where $v_b$ is the speed of the belt and $\theta'$ is the angle of motion relative to the belt.

 

[IamVector]

$\frac 2 {\sqrt 5} m/s$ is the answer I get.

In general the minimum speed is $v_b \cos \theta'$, where $v_b$ is the speed of the belt and $\theta'$ is the angle of motion relative to the belt.

can you please explain further will be a great help

 

[IamVector]

$\frac 2 {\sqrt 5} m/s$ is the answer I get.

In general the minimum speed is $v_b \cos \theta'$, where $v_b$ is the speed of the belt and $\theta'$ is the angle of motion relative to the belt.

is my approach right just want to confirm??

 

[PeroK]

can you please explain further will be a great help

That's what comes out of the algebra and calculus. I suspect the geometric approach could be generalised.

 

[IamVector]

That's what comes out of the algebra and calculus. I suspect the geometric approach could be generalised.

is my approach correct?? otherwise I will try it with different approach cause sometimes answer is correct approach is wrong

 

[PeroK]

is my approach correct?? otherwise I will try it with different approach cause sometimes answer is correct approach is wrong

I haven't seen your approach.

 

[IamVector]

I haven't seen your approach.

ok , so i got this equation (-1+μ gt/(5^1/2))i + (2-2μ gt/(5^1/2))j so to convert it into lab frame or ground frame I added 1m/s ie velocity of belt then I took the mod of the resultant vector ie (-1+μ gt/(5^1/2))i + (2-2μ gt/(5^1/2))j which in return formed a quadratic in μgt now we were asked to find the minimum vector so it will be minimum when the value of quadratic will be minimum so by differentiating i found the min val to be 4/5 thus on sqrt gives 2/(5^1/2)

 

[IamVector]

I haven't seen your approach.

I am preparing for Olympiads a word of advice will be appreciated and are the level of difficulty of questions I am doing sufficient??

 

[PeroK]

I am preparing for Olympiads a word of advice will be appreciated and are the level of difficulty of questions I am doing sufficient??

I don't know. Try this one:

https://www.physicsforums.com/threads/find-initial-acceleration-in-a-system.985009/

Don't take all my advice on that thread! Looking at energy wasn't much use. Forces and torques.

 

[IamVector]

I don't know. Try this one:

https://www.physicsforums.com/threads/find-initial-acceleration-in-a-system.985009/

Don't take all my advice on that thread! Looking at energy wasn't much use. Forces and torques.

Is it g/2? I am not sure

 

[PeroK]

Is it g/2? I am not sure

That's a quite a hard problem I think. You can't just guess that the answer is $g/2$ or $3g/13$ or whatever. You have to work it out.