Hard kinematics problem: Block pushed onto a moving conveyor belt

[IamVector]

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

 

[PeroK]

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?

 

[Lnewqban]

Welcome, IamVector!
It seems that this problem is a subtraction of vectors.
Could you post a diagram?

 

[IamVector]

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

 

[PeroK]

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!

 

[IamVector]

Okay, let the coefficient of friction be $\mu$. Now it is given!

ok I will try the best I can

 

[IamVector]

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??

 

[PeroK]

do I have to represent k in terms of friction coefficient and time??

You need the equatiion of motion in the belt frame, yes.

 

[IamVector]

You need the equatiion of motion in the belt frame, yes.

W = ((5^1/2)-µgt)?? so k = (1-((µgt)/5^1/2)

 

[IamVector]

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

 

[PeroK]

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?

 

[IamVector]

Forget about the ground frame for the moment. Can you give the components of the velocity in the belt frame?

-1i+2j ??

 

[PeroK]

-1i+2j ??

Come on! As a function of time.

 

[IamVector]

Come on! As a function of time.

-(1-µgt)i + ( 2-µgt)j ?

 

[PeroK]

-(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.

 

[jbriggs444]

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.

-(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.

 

[IamVector]

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

 

[jbriggs444]

(-(1)i + ( 2)j) -μgt

The difficulty now is that you do not have a direction on the $\mu g t$ vector.

 

[IamVector]

The difficulty now is that you do not have a direction on the $\mu g t$ vector.

will it be unit vector * friction force??

 

[jbriggs444]

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.

 

[IamVector]

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??

 

[jbriggs444]

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.

 

[IamVector]

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??

 

[jbriggs444]

So we have a

(-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?

 

[IamVector]

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??

 

[jbriggs444]

(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.

 

[PeroK]

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.

 

[IamVector]

Yes, that looks right to me. However, you've lost the initial velocity components.

(-1+μgt/√5 )i-(2-2μgt/√5)j??

 

[jbriggs444]

(-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.

 

[IamVector]

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??