Sphere Slipping Against Blocks: Solving for the Speed of the Sphere's Center

In summary: I am really very very sorry for the inconvenience caused .I have even mailed the site admin to delete the question .Perhaps the question is too ill defined to be of any use to anybody .Once again I am sorry .In summary, the given arrangement consists of a sphere of radius R placed on two blocks, A and B. Block A is stationary while block B is moving at a constant speed v towards the left. The goal is to find the speed of the sphere's center when the distance between the blocks is √2R and the sphere's center is at a vertical distance R/√2 from the top of the blocks. After solving the problem, it is determined that the speed of the sphere's center is v
  • #1
Tanya Sharma
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Sphere slipping against blocks

Homework Statement



In the given arrangement ,a sphere of radius R is placed on the two blocks A and B where A is fixed and B is moving at a constant speed ‘v’ towards left .Find the speed of the sphere’s center when the distance between the blocks is √R and the sphere’s center is at a vertical distance √R from the top of the blocks .All surfaces are frictionless .

Ans : v/2

Homework Equations





The Attempt at a Solution



Let the distance between the blocks be 'y' at any instant.
Let the distance between the sphere’s center be at a vertical distance 'x' from the top of the blocks .
Let the horizontal distance between B and sphere's center O be 'l' .

Then l=y/2 .

Differentiating we get,dl/dt = (1/2)dy/dt = v/2

From the geometry of the setup , x2 + l2 = R2

Differentiating , we get dx/dt = -(l/x)dl/dt .

Now when x=√R and y=√R (i.e l=√R/2) ,we have dx/dt = -v/4 i.e the horizontal speed of O is v/4 downwards .

The center also has a horizontal speed v/2 towards left.

The net speed of the center is (√5/4)v which is incorrect .

I would be grateful if someone could help me with the problem .
 

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  • #2
Hmmm, maybe I'm not understanding the problem correctly, but the geometry for this problem seems to be over-constrained.

From the problem statement:

  • The distance from the center of the sphere to the contact point on block B (or A for that matter) is (√R)/2. We'll call this B.
  • The distance from the center of the sphere to the top of the blocks is √R. We'll call this C.
  • This distance from the contact point on block B to the center of the sphere is the radius R. We'll call this A.

These three constraints form a right triangle, and the Pythagorean theorem tells us that A = √(B^2+C^2). Plugging the above in gives us:

R = √(((√R)/2)^2+(√R)^2) = √((R/4)+√(R)) = (√5/4)R

This shows that the given dimensions cannot result in a right triangle, suggesting an ill-defined problem.
 
  • #3
Hello, Tanya.

I'm wondering if there's a mistake in the wording of the problem. It doesn't seem dimensionally correct to take the distance between the blocks to be √R. Anyway, if you let x = √R and l = (√R)/2, then you certainly wouldn't satisfy x2 + l2 = R2.

[EDIT: Incorrect statement deleted.]
 
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  • #4
I tried working backwards from the v/2 answer to figure out the misprint. I deduced that the blocks are touching! And if you think about that arrangement you can see it does indeed give v/2. The centre of the sphere will be moving left at half the speed of the block.
Before I did that, I would have guessed it was supposed to be a separation of R√2, with the sphere's centre initially R/√2 above the blocks. That gives a speed of v/√2. Could this be a record for misprints in one question?

Tanya, to get some benefit from the question, try taking the separation to be some unknown x and post what you get.
 
  • #5
Hi TSny…

It seems the new year hasn’t started well as far as putting up a question on PF is concerned.I regret putting up an ill defined problem :redface:.

Anyways I feel I have figured out the problem with the problem .Interestingly,all the radical signs have gone topsy turvy .

Here is what I believe the question should have been :

"In the given arrangement ,a sphere of radius R is placed on the two blocks A and B where A is fixed and B is moving at a constant speed ‘v’ towards left .Find the speed of the sphere’s center when the distance between the blocks is √2R and the sphere’s center is at a vertical distance R/√2 from the top of the blocks .All surfaces are frictionless .

Ans:v/√2 "


What is your opinion ?
 
  • #6
haruspex said:
Before I did that, I would have guessed it was supposed to be a separation of R√2, with the sphere's centre initially R/√2 above the blocks. That gives a speed of v/√2.

I feel the same .

haruspex said:
Could this be a record for misprints in one question?

Yes... But this record has left me :redface:.
 
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  • #7
Tanya Sharma said:
Hi TSny…
Here is what I believe the question should have been :

"In the given arrangement ,a sphere of radius R is placed on the two blocks A and B where A is fixed and B is moving at a constant speed ‘v’ towards left .Find the speed of the sphere’s center when the distance between the blocks is √2R and the sphere’s center is at a vertical distance R/√2 from the top of the blocks .All surfaces are frictionless .

Ans:v/√2 "


What is your opinion ?

I like it. It's the same as haruspex's prognosis!
 
  • #8
Thanks TSny...

Actually I figured out the problem with the problem the moment I saw the initial response .But found it quite hard to believe that the question had so many mistakes .I even responded just after you have replied , only to delete the post .
 

1. What is the difference between rolling and slipping?

Rolling refers to the motion of a round object, such as a sphere, where it rotates around a fixed point without sliding. Slipping, on the other hand, occurs when the sphere's motion includes both rotation and sliding along a surface.

2. How does friction play a role in sphere slipping against a block?

Friction is the force that resists the sliding motion between two surfaces. In the case of a sphere slipping against a block, friction is responsible for reducing the speed of the sphere's rotation and causing it to slip along the surface of the block.

3. Can the coefficient of friction affect the slipping of a sphere against a block?

Yes, the coefficient of friction, which is a measure of the roughness of the surfaces in contact, can affect the slipping of a sphere against a block. A higher coefficient of friction means there is more resistance to motion, resulting in a slower slipping motion of the sphere.

4. What factors determine whether a sphere will roll or slip against a block?

The main factors that determine whether a sphere will roll or slip against a block are the shape and roughness of the surfaces in contact, the speed and direction of the sphere's motion, and the force of friction between the surfaces.

5. How does the weight of the sphere affect its slipping motion against a block?

The weight of the sphere, or its mass, plays a role in its slipping motion against a block. A heavier sphere will have more inertia and be more resistant to changes in its motion, resulting in a slower slipping motion compared to a lighter sphere.

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