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14. Moving cylinder

Place a sheet of paper on a horizontal table and put a cylindrical object (e.g. a pencil) on the paper.

Pull the paper out. Observe and investigate the motion of the cylinder until it comes to rest.

My name is Zadayanchuk Andrey. I would like to present you the solution of problem No 14 “Moving cylinder”

3. Here is a plan of my report. First of all we will analyze movement of a cylinder on a sheet of paper when we pull the sheet out of the cylinder. Then we will analyze movement of a rotating cylinder across a horizontal table. Our aim is to find distances traveled by cylinder though the paper and though the table. In the end of my report a conclusion will be made.

Stage 1

4. Let’s suppose that sheet of paper is moving under the cylinder with acceleration as. Our first task is to find acceleration of cylinder’s axis (a). To do this we use the second Newton’s law for translational and rotational motions of the cylinder (formula 1 and 2). Here Ffr is the friction force between cylinder and paper , M is the moment of friction (see formula 3). And (4) is a coupling equation for angular acceleration of cylinder and tangential acceleration of cylinder’s surface relatively to cylinder’s axis (atau). Here ksi is dimensionless parameter depending on mass distribution in the cylinder (for thin-walled cylinder ksi is equal to 1, for solid cylinder ksi is equal to one half).

5.

If there is no sliding between surfaces of paper and cylinder the axis’ acceleration can be calculated using formula (5). In this case using formula 1-5 gives us formula 6 from which simple mathematical transformation yields formula (7 )

6. Calculation of critical acceleration (a_sc)

Critical acceleration is an acceleration of paper when sliding between surfaces of sheet and cylinder occurs. It means that a friction force is greater than maximal value of static friction force (formula 8). It means that maximal acceleration of cylinder’s axis is given by formula 9. Using formula 7 we get formula 10.

7. Acceleration of cylinder’s axis

Thus acceleration of axis can be calculated using combined formula 11. The upper part of this formula is for case of static friction, the lower part of it is for case of sliding friction.

Now, when we know acceleration of cylinder we can find displacement of cylinder and displacement of sheet of paper while cylinder and paper are in contact.

8. Relation displacement with acceleration

Let time of pulling out of sheet of paper from under the cylinder is t. In this case displacement of cylinder axis S1 is given by formula 12. Displacement of sheet of paper itself is given by formula 13. In addition displacement of sheet of paper is equal to the sum of displacement of cylinder’s axis and displacement of paper relatively to the cylinder (formula 14). Dividing of formula 12 by formula 13 gives formula 15. Since we know expressions for both accelerations in this formula we can find displacement of the cylinder while the paper is pulled out from under it. Remember that there are two different cases of pulling out the paper.

9. Pulling out without sliding

When sliding is absent, we have got formula 7 for cylinder’s acceleration. Substitution of it into formula 15 gives formula 16 from which it is easy to get final formula for cylinder’s displacement S1 (formula 17). Recall that l horizontal distance from the initial position of the cylinder and the edge of the sheet of paper.

10. Pulling out with sliding

Place a sheet of paper on a horizontal table and put a cylindrical object (e.g. a pencil) on the paper.

Pull the paper out. Observe and investigate the motion of the cylinder until it comes to rest.

My name is Zadayanchuk Andrey. I would like to present you the solution of problem No 14 “Moving cylinder”

3. Here is a plan of my report. First of all we will analyze movement of a cylinder on a sheet of paper when we pull the sheet out of the cylinder. Then we will analyze movement of a rotating cylinder across a horizontal table. Our aim is to find distances traveled by cylinder though the paper and though the table. In the end of my report a conclusion will be made.

Stage 1

4. Let’s suppose that sheet of paper is moving under the cylinder with acceleration as. Our first task is to find acceleration of cylinder’s axis (a). To do this we use the second Newton’s law for translational and rotational motions of the cylinder (formula 1 and 2). Here Ffr is the friction force between cylinder and paper , M is the moment of friction (see formula 3). And (4) is a coupling equation for angular acceleration of cylinder and tangential acceleration of cylinder’s surface relatively to cylinder’s axis (atau). Here ksi is dimensionless parameter depending on mass distribution in the cylinder (for thin-walled cylinder ksi is equal to 1, for solid cylinder ksi is equal to one half).

5.

If there is no sliding between surfaces of paper and cylinder the axis’ acceleration can be calculated using formula (5). In this case using formula 1-5 gives us formula 6 from which simple mathematical transformation yields formula (7 )

6. Calculation of critical acceleration (a_sc)

Critical acceleration is an acceleration of paper when sliding between surfaces of sheet and cylinder occurs. It means that a friction force is greater than maximal value of static friction force (formula 8). It means that maximal acceleration of cylinder’s axis is given by formula 9. Using formula 7 we get formula 10.

7. Acceleration of cylinder’s axis

Thus acceleration of axis can be calculated using combined formula 11. The upper part of this formula is for case of static friction, the lower part of it is for case of sliding friction.

Now, when we know acceleration of cylinder we can find displacement of cylinder and displacement of sheet of paper while cylinder and paper are in contact.

8. Relation displacement with acceleration

Let time of pulling out of sheet of paper from under the cylinder is t. In this case displacement of cylinder axis S1 is given by formula 12. Displacement of sheet of paper itself is given by formula 13. In addition displacement of sheet of paper is equal to the sum of displacement of cylinder’s axis and displacement of paper relatively to the cylinder (formula 14). Dividing of formula 12 by formula 13 gives formula 15. Since we know expressions for both accelerations in this formula we can find displacement of the cylinder while the paper is pulled out from under it. Remember that there are two different cases of pulling out the paper.

9. Pulling out without sliding

When sliding is absent, we have got formula 7 for cylinder’s acceleration. Substitution of it into formula 15 gives formula 16 from which it is easy to get final formula for cylinder’s displacement S1 (formula 17). Recall that l horizontal distance from the initial position of the cylinder and the edge of the sheet of paper.

10. Pulling out with sliding

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