Solve Projected Cylinder: Find Distance s w/ 3 Solutions

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In summary, there is a discrepancy between the two solutions for the distance traveled by the center of a solid uniform cylinder rolling without slipping up an incline. One solution neglects the effect of rolling and suggests s=V๐²/4gsinӨ, while the other solution takes into account the rolling and suggests s=3V๐²/4gsinӨ. It is recommended to check the calculations and possibly contact the authors for clarification.
  • #1
yasar1967
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1. A solid uniform cylinder of mass M and Radius is projected up an incline of angle Ө. It rolls without slipping from an initial speed V๐ of the center of mass. What distance s does the center of cylinder travel before it starts to fall back?
2. I=½MR², V²=V๐²+2as, F=ma, a=R²α ...
3. MgsinӨ+f=ma=mR²α , f=Iα/R

0²=V๐²-2as , I=I=½MR² , a=2gsinӨ
s=V๐²/4gsinӨ
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But the answer page of the book says s= (3V๐²/4g) sinӨ [exactly as it's written---Prob. 9/67 PHYSICS for... 3rd Edition; Fishbane-Gasiorowich-Thornton]
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I've found another solution from another book which goes with the conservation of the energy:
½MV๐²+½Iω๐²=MgH , H=SsinӨ
For a solid uniform cylinder, rotational kinetic energy is 1/3th of total energy and therefore:
3/2(½MV๐²)=MgSsinӨ, s=3V๐²/4gsinӨ

Which of us is correct?
 
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I'm surprised that one one yet has an answer.
Maybe it's a good idea to get in touch with the authors 'cos I believe they made a mistake either in the book's second or third edition.
 
  • #3
yasar1967 said:
0²=V๐²-2as , I=I=½MR² , a=2gsinӨ
s=V๐²/4gsinӨ

You should not use this because this neglects the affect of the rolling of the cylinder. Also, check your acceleration with a force-diagram (unless you meant SsinӨ).

However, as you have already stated
yasar1967 said:
½MV๐²+½Iω๐²=MgH , H=SsinӨ
, the conservation of energy includes the rolling. Make sure to check your calculations.

---
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What is a projected cylinder?

A projected cylinder is a geometric shape that is created by extending a circle along an axis in a straight line, resulting in a 3-dimensional shape that resembles a cylinder. It is commonly used in engineering and mathematics to model various real-world objects.

What is the purpose of solving a projected cylinder?

The purpose of solving a projected cylinder is to find the distance between two points on the surface of the cylinder. This can be useful in various applications, such as calculating the length of a pipe or the volume of a cylinder-shaped container.

How do you find the distance s in a projected cylinder?

To find the distance s in a projected cylinder, you will need to use the Pythagorean theorem, which states that the square of the hypotenuse (longest side) of a right triangle is equal to the sum of the squares of the other two sides. In this case, the hypotenuse represents the distance s, and the other two sides represent the known dimensions of the cylinder.

Why are there 3 solutions for finding the distance s in a projected cylinder?

There are 3 solutions for finding the distance s in a projected cylinder because the Pythagorean theorem can be applied in three different ways depending on the given information. For example, if you know the radius and height of the cylinder, you can use the formula s = √(r² + h²) to find the distance. But if you only know the radius and the distance from the center to the desired point on the surface, you would use the formula s = √(2r² - d²).

What are some common real-world examples of solving a projected cylinder?

Solving a projected cylinder can be used in various real-world examples, such as finding the length of a pipe or the volume of a cylindrical tank. It can also be used in engineering to calculate the distance between two points on a curved surface, such as the inside of a pipe or the outside of a cylinder-shaped object.

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