# Newton's laws--Block sliding up a frictionless semi-circular track

• sachin
In summary: The equation of motion for the small block is:##u## is the velocity of the small block of mass ##m##.## r ## is the radius of the semicircle.## \theta ## is the angle of the mass ##m## w.r.t vertical.## \frac{m}{r} \dot u - Nm \dot r = 0$$where:##N## is the total mass of the two blocks##mg## is the mass of the big block##m## is the mass of the small block sachin Homework Statement velocity relation in constrained motion Relevant Equations momentum conservation While coming across a question on Newton's laws,I found a case when a block slides on another block kept horizontally as in the figure.All surfaces are frictionless. My concern is for the blocks to be in contact that is to move together will the horizontal velocities of both the blocks be the same in every part of the semicircular path or it happens only at the point A. #### Attachments • IMG_20220817_184248 (1).JPG 13.5 KB · Views: 82 Welcome to PF. sachin said: Homework Statement:: velocity relation in constrained motion Relevant Equations:: momentum conservation My concern is for the blocks to be in contact that is to move together will the horizontal velocities of both the blocks be the same in every part of the semicircular path or it happens only at the point A. Sorry, but I'm having trouble figuring out your problem statement. Are you trying to figure out when the block will lose contact with the vertical semi-circular part of the track? The horizontal velocities are only necessarily equal at point A. have posted the whole question in the figure. #### Attachments • IMG_20220817_020550.JPG 42.1 KB · Views: 75 Dale said: The horizontal velocities are only necessarily equal at point A. my concern is why only at A. berkeman said: Welcome to PF.Sorry, but I'm having trouble figuring out your problem statement. Are you trying to figure out when the block will lose contact with the vertical semi-circular part of the track? I am asking for any point on the curve,what is the relation between the velocities of the blocks to be in contact i.e to move together. sachin said: my concern is why only at A. A is the only point where the normal is horizontal. The big block acts as a constraint, the small block cannot go into the big block, which means that the component of the velocity of the small block in the normal direction cannot be smaller than the component of the velocity of the large block in the normal direction (where the normal direction is defined to point outside the large block) but there is no velocity of the smaller block along he normal as its always pependicular to the common normal. sachin said: but there is no velocity of the smaller block along he normal as its always pependicular to the common normal. That is only true if: 1) the small block remains in contact with the large block and 2) the large block is stationary in the curve,when seen in the frame of the bigger block,the smaller block always moves tangentially i.e perpenduclar to the normal,if the horizontal velocitie are not same then the blocks will get detached and wil move idependently,so everywhere in the motion,they are in contact as are moving together. the question is discussed in the below youtube link, sachin said: when seen in the frame of the bigger block That is what is implied by my point (2) above. However, even in that frame it is still important to consider the possibility that the small block does not remain in contact. E.g. when the small block passes point A then it is possible for gravity to pull it off the big block. sachin said: if the horizontal velocitie are not same then the blocks will get detached and wil move idependently Yes, indeed. That is why (1) is not always true. I am not sure what more you are looking for. I gave the general rule in post 7. You gave a special case rule in post 8. I explained the conditions for the special case to be valid in post 9. And now we are just re-discussing those conditions. Do you have anything else to discuss? Lnewqban sachin said: Homework Statement:: velocity relation in constrained motion Relevant Equations:: momentum conservation While coming across a question on Newton's laws,I found a case when a block slides on another block kept horizontally as in the figure.All surfaces are frictionless. My concern is for the blocks to be in contact that is to move together will the horizontal velocities of both the blocks be the same in every part of the semicircular path or it happens only at the point A. Assuming the small block has been given enough initial velocity to get it to A in the first place, how could its velocity be the same as the big block at any other point ? Last edited: sachin said: in the frame of the bigger block, the smaller block always moves tangentially i.e perpenduclar to the normal, Yes. sachin said: if the horizontal velocitie are not same then the blocks will get detached No, in general it means their velocities along the normal are the same. Only at point A is the normal horizontal. Before it reaches point A, the small block is moving left faster than the large block. Lnewqban Out of curiosity, for the equations of motion at the semicircle: ##u## is the velocity of the large block of mass ##M##. ## r ## is the radius of the semicircle. ## \theta ## is the angle of the mass ##m## w.r.t vertical. ## \nwarrow_{\theta}^+ \nearrow_r^+ ##$$ \begin{array} \, N - mg \cos \theta + m \dot u \sin \theta = mr {\dot \theta}^2 \\ \quad \\ N \sin \theta = M \dot u \\ \quad \\ \dot u \cos \theta - g \sin \theta =r \ddot \theta \end{array} 

Does that system equations look correct?

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hmmm27 said:
Assuming the small block has enough initial velocity to get it to A in the first place, how could its velocity be the same as the big block at any other point ?
if the horizontal velocities are not the same at every point,how blocks remain in contact.

sachin said:
if the horizontal velocities are not the same at every point,how blocks remain in contact.

If the vx were the same, the smaller block couldn't move in x-direction in respect to the larger block. But the smaller block does indeed move in x-derection when following the circular curve.I hope that makes sense to you.

sachin said:
if the horizontal velocities are not the same at every point,how blocks remain in contact
The small block slides along the surface of the larger block, perpendicular to the normal which is not horizontal at every point.

Ahzmandius said:
In your first equotion you add a y component of the accleration to the x component of that acceleration.

These are ##r,\theta## components, I'm not working in rectangular coordinates?

erobz said:
These are ##r,\theta## components, I'm not working in rectangular coordinates?
Yeah I know it was a draft that I forget to delete. I did now.

Imagine the mass ##m## being held still at some angle ##\theta## traveling with the mass ##M##. Masses ##M## and ##m## have the same velocity in that case. Now, if the small mass ##m## has some angular velocity at some ##\theta## (which it does) then it has some tangential velocity at ##\theta##. A component of that tangential velocity will be in the direction of the motion of the large blocks velocity. Meaning its velocity in the horizontal direction (as seen from the ground) will be greater than the velocity of the large mass. Point A is a special place, because the component of mass ##m##'s tangential velocity in the horizontal direction is zero.

sachin said:
my concern is why only at A.
Once the small block moves up passing point A, it does not have any more influence on the leftward movement of the big block.
The small block is unable to pull the big block, only to push it.

This situation could be compared to a perfectly elastic collision: beyond point A, it is like the small block is bouncing back from the vertical wall of the big block.

Lnewqban said:
Once the small block moves up passing point A, it does not have any more influence on the leftward movement of the big block.
Yes, it does. As long as the normal force is non-zero and the normal direction is not vertical, it has an influence.

Lnewqban said:
The small block is unable to pull the big block, only to push it.
It can still push even beyond A, as long as it has not yet been pulled free (or sped free) from contact.

for the blocks to be in contact their velocities along the common normal have to be the same but I am unable to get the expression of velocities.

sachin said:
for the blocks to be in contact their velocities along the common normal have to be the same but I am unable to get the expression of velocities.
The velocity of ##M## is only in the horizontal direction. You mean the component of their velocity in the normal direction must be equal I believe.

sachin said:
for the blocks to be in contact their velocities along the common normal have to be the same but I am unable to get the expression of velocities.
Have you considered conservation of mechanical energy? Surely there is an equation to be had there?

Also, telling us what you cannot do is not nearly as helpful as telling us what you have tried.

erobz
erobz said:
The velocity of ##M## is only in the horizontal direction. You mean the component of their velocity in the normal direction must be equal I believe.
yes,at any point in the curve the velocitis in normal directions must be the same, else they l loose contact.

jbriggs444 said:
Have you considered conservation of mechanical energy? Surely there is an equation to be had there?

Also, telling us what you cannot do is not nearly as helpful as telling us what you have tried.
i believe,if the x velocity(horizontal) of the bigger block is v,then x velocity of the smaller on will be u cos thita + v,where u is the relative velocity of the smaller block wrt the bigger one.

sachin said:
i believe,if the x velocity(horizontal) of the bigger block is v,then x velocity of the smaller on will be u cos thita + v,where u is the relative velocity of the smaller block wrt the bigger one.
That sounds a bit like conservation of momentum to me. But that cannot be right since the masses of the two blocks do not appear anywhere in the equation. Then too, the letter "v" is already taken. It denotes the velocity of the small block. See your post #1 above.

jbriggs444 said:
That sounds a bit like conservation of momentum to me. But that cannot be right since the masses of the two blocks do not appear anywhere in the equation. Then too, the letter "v" is already taken. It denotes the velocity of the small block. See your post #1 above.

my concern from the beginning is clear,I just want to know if the x velocities are not the same in the curve,how the bodies dnt loose contact or what velocities are equal for the contact to be maintained,have searched that the components along the normal should be the same but the smaller block does not have any component along the normal in the
curve.

sachin said:
just want to know if the x velocities are not the same in the curve
They are not, except at point A.

sachin said:
how the bodies dnt loose contact or what velocities are equal for the contact to be maintained
The component of the velocities in the normal direction must be equal to maintain contact

So is everything resolved? If not, what remains? Please be as clear as possible

sachin said:
I just want to know if the x velocities are not the same in the curve
It is far from clear. But I think that you are confusing velocity with speed, invoking conservation of momentum, adopting a center of momentum frame and forgetting that the two blocks may have different masses.

If the two blocks have equal masses and start with zero total momentum then conservation of momentum immediately demands that the horizontal component of their velocities are at all times equal and opposite so that their horizontal speeds are at all times equal.

if the x velocities are different,why does not the small block pierce through and enter inside the bigger block.

sachin said:
if the x velocities are different,why does not the small block pierce through and enter inside the bigger block.
Because the direction “inside the bigger block” is the normal direction, not the x direction

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Lnewqban

W.r.t the ground the velocity of the sled when the mass is at some angle ##\theta## is represented by the green arrow

W.r.t the ground the velocity of the mass on the sled in the horizontal direction is the projection of the red arrow onto the horizontal direction plus the green arrow.

In general, the velocity of the sled in the horizontal direction will not equal the velocity of the mass in the horizontal direction.

It is at point A, when the projection of the red arrow onto the green arrow is ##0##, that the velocity of the sled matches the velocity of the mass.

Lnewqban

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