Hard kinematics problem: Block pushed onto a moving conveyor belt

In summary: I think you have a sign error there. A component of the frictional force should have the opposite sign to the corresponding component of the...
  • #1
IamVector
98
9
Homework Statement
A block is pushed onto a conveyor belt. The belt
is moving at velocity v0 = 1 m/s, the block’s initial velocity
u0 = 2 m/s is perpendicular to the belt’s velocity. During its
subsequent motion, what is the minimum velocity of the block
with respect to the ground? The coefficient of friction is large
enough to prevent the block from falling off the belt.
Relevant Equations
hint given : we use the conveyor’s frame, but
as we are asked about the speed in lab frame, we need
to switch back to the lab frame. In the conveyor’s frame,
the velocity vector becomes shorter while preserving the
direction, i.e. can be represented as ⃗w = k ⃗w0, where its
initial value ⃗w0 = ⃗v0 −⃗u0 and the factor k takes numerical
values from 0 to 1. Hence, the velocity in the lab frame
⃗v = ⃗u0 + k ⃗w0: this is a vector connecting the right angle
of the right triangle defined by its catheti ⃗u0 and ⃗v0 with a
point on the hypotenuse; the specific position of this point
depends on the value of the factor k (which is a function
of time).
stuck on this question
 
Physics news on Phys.org
  • #2
If you are not given much information, then you have to use the little you are given. There must be some finite coefficient of friction. What can you do with that?
 
  • Like
Likes IamVector
  • #3
Welcome, IamVector!
It seems that this problem is a subtraction of vectors.
Could you post a diagram?
 
  • #4
PeroK said:
If you are not given much information, then you have to use the little you are given. There must be some finite coefficient of friction. What can you do with that?
there is no finite coefficient of friction given does it have any relation with the velocities given ?? it is just stated that it is large enough to prevent it from falling off the belt
 
  • #5
IamVector said:
there is no finite coefficient of friction given does it have any relation with the velocities given ?? it is just stated that it is large enough to prevent it from falling off the belt
Okay, let the coefficient of friction be ##\mu##. Now it is given!
 
  • #6
PeroK said:
Okay, let the coefficient of friction be ##\mu##. Now it is given!
ok I will try the best I can
 
  • #7
PeroK said:
Okay, let the coefficient of friction be ##\mu##. Now it is given!
do I have to represent k in terms of friction coefficient and time??
 
  • #8
IamVector said:
do I have to represent k in terms of friction coefficient and time??
You need the equatiion of motion in the belt frame, yes.
 
  • #9
PeroK said:
You need the equatiion of motion in the belt frame, yes.
W = ((5^1/2)-µgt)?? so k = (1-((µgt)/5^1/2)
 
  • #10
PeroK said:
You need the equatiion of motion in the belt frame, yes.
can you please explain me the hint given?it will be a great help
 
  • #11
IamVector said:
can you please explain me the hint given?it will be a great help
Forget about the ground frame for the moment. Can you give the components of the velocity in the belt frame?
 
  • #12
PeroK said:
Forget about the ground frame for the moment. Can you give the components of the velocity in the belt frame?
-1i+2j ??
 
  • #13
IamVector said:
-1i+2j ??
Come on! As a function of time.
 
  • #14
PeroK said:
Come on! As a function of time.
-(1-µgt)i + ( 2-µgt)j ?
 
  • #15
IamVector said:
-(1-µgt)i + ( 2-µgt)j ?
Friction doesn't know about ##x## and ##y## directions. It opposes motion uniformly. You may need to think about that.
 
  • #16
Trying a slightly different approach...

You can follow the @PeroK approach and work this algebraicly. Or you can dispense with the numbers and coordinates for the moment and deal with it geometrically.

IamVector said:
-(1-µgt)i + ( 2-µgt)j ?
The decelleration has a total magnitude of ##\mu g##. Your proposed formula puts the angle of the frictional force at 45 degrees with total magnitude ##\sqrt{2} \mu g##. That is not right. You need for the two component forces to make the correct angle and the correct magnitude.

Proceeding with the geometric approach.

From the belt frame, you have this object arriving, at -1i + 2j just as you state. The only forces on the object are gravity, normal force and friction. Gravity and normal force cancel. It will follow a straight-line path at constant deceleration until it comes to a stop.
 
  • Like
Likes PeroK
  • #17
PeroK said:
Friction doesn't know about ##x## and ##y## directions. It opposes motion uniformly. You may need to think about that.
(-(1)i + ( 2)j) -μgt
 
  • #18
IamVector said:
(-(1)i + ( 2)j) -μgt
The difficulty now is that you do not have a direction on the ##\mu g t## vector.
 
  • #19
jbriggs444 said:
The difficulty now is that you do not have a direction on the ##\mu g t## vector.
will it be unit vector * friction force??
 
  • #20
IamVector said:
will it be unit vector * friction force??
If you want to keep working with components (as @PeroK surely intends) then you need to split that ##-\mu g t## into ##\hat{i}## and ##\hat{j}## components.

Time to apply some trig functions.

With the geometric approach, I would have you draw that (-1,2) velocity vector and imagine that you are searching for some lesser velocity vector along the same line.

The lesser velocity vector you are searching for is the one that you can add (vector addition) to the belt speed to minimize the magnitude of the resulting vector sum.
 
  • #21
jbriggs444 said:
If you want to keep working with components (as @PeroK surely intends) then you need to split that ##-\mu g t## into ##\hat{i}## and ##\hat{j}## components.

Time to apply some trig functions.
so angle between resultant and i component will be arctan(2/1) so I used it to find the components of friction force
-(-μgt/√5 )i+(-2μgt/√5)j??
 
  • #22
IamVector said:
so angle between resultant and i component will be arctan(2/1) so I used it to find the components of friction force
-(-μgt/√5 )i+(-2μgt/√5)j??
I think you have a sign error there. A component of the frictional force should have the opposite sign to the corresponding component of the velocity.
 
  • #23
jbriggs444 said:
I think you have a sign error there. A component of the frictional force should have the opposite sign to the corresponding component of the velocity.
(-μgt/√5 )i-(-2μgt/√5)j??
 
  • #24
So we have a
IamVector said:
(-μgt/√5 )i-(-2μgt/√5)j??
Ummm. Check the signs on your velocity components against the signs on your acceleration components. Does friction cause acceleration or decelleration?
 
  • #25
jbriggs444 said:
So we have a

Ummm. Check the signs on your velocity components against the signs on your acceleration components. Does friction cause acceleration or decelleration?
(μgt/√5 )i-(2μgt/√5)j??
 
  • #26
IamVector said:
(μgt/√5 )i-(2μgt/√5)j??
Yes, that looks right to me. However, you've lost the initial velocity components.

Or, perhaps, by now you've forgotten what you were setting out to calculate. We want a formula for component-wise velocity that starts out high at t=0 and ends up low some time later.
 
  • #27
jbriggs444 said:
If you want to keep working with components (as @PeroK surely intends) then you need to split that ##-\mu g t## into ##\hat{i}## and ##\hat{j}## components.

Time to apply some trig functions.

With the geometric approach, I would have you draw that (-1,2) velocity vector and imagine that you are searching for some lesser velocity vector along the same line.

The lesser velocity vector you are searching for is the one that you can add (vector addition) to the belt speed to minimize the magnitude of the resulting vector sum.
Just to explain the approach I was thinking about. In the belt frame, for which I'll use primed values.

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:
$$v'_x = -v'\sin \theta', \ \ v'_y = v'\cos \theta'$$
And now we can transform to the ground frame ...

Hint: ##v## is minimised when ##v^2## is minimised.
 
  • #28
jbriggs444 said:
Yes, that looks right to me. However, you've lost the initial velocity components.
(-1+μgt/√5 )i-(2-2μgt/√5)j??
 
  • Like
Likes jbriggs444
  • #29
IamVector said:
(-1+μgt/√5 )i-(2-2μgt/√5)j??
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.
 
  • #30
PeroK said:
Just to explain the approach I was thinking about. In the belt frame, for which I'll use primed values.

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:
$$v'_x = -v'\sin \theta', \ \ v'_y = v'\cos \theta'$$
And now we can transform to the ground frame ...

Hint: ##v## is minimised when ##v^2## is minimised.
initial angle between belt velocity and block velocity??
 
  • #31
so now how do we convert into lab frame?? by using hint but what is the logic behind it??
 
  • #32
IamVector said:
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.
 
  • #33
jbriggs444 said:
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??
 
  • #34
PeroK said:
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:
IamVector said:
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.
 
  • Like
Likes PeroK
  • #35
jbriggs444 said:
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 .
 

Similar threads

  • Introductory Physics Homework Help
Replies
11
Views
817
  • Introductory Physics Homework Help
Replies
8
Views
1K
  • Introductory Physics Homework Help
Replies
8
Views
2K
  • Introductory Physics Homework Help
3
Replies
78
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
895
  • Introductory Physics Homework Help
Replies
6
Views
7K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
25
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
2
Replies
58
Views
3K
Back
Top