Another Solution check Please. (cylinder rolling down a slope)

In summary, the cylindrical hoop's final height after rolling without slipping down a rough incline and then sliding up a perfectly smooth incline can be found using the conservation of energy equation, where the final velocity is equal to the square root of the initial height times gravity. As it slides up the incline, the hoop loses the ability to roll without slipping due to the lack of friction, and the rotational and translational motions become independent. This results in the rotational energy remaining constant while the translational energy decreases.
  • #1
phalanx123
30
0
Sorry here is another question that I am not sure of

1. A cylindrical hoop rests on a rough uniform incline. It is released and rolls
without slipping through a vertical distance h0. It then continues up a perfectly
smooth incline. What height does it reach?



Here is my solution

Because of the conservasion of energy

mgh0=1/2mv^2+1/2IOmega^2

since Omega=v/r and the moment of inertia of a cylindrical hoop is I=mr^2

therefore

mgh0=1/2mv^2+1/2mr^2*(v/r)^2
=mv^2

therefore final velocity is v=Square root(gh0)

when goin up the perfect smooth surface it looses the ability to roll, so the only energy is the translaional energy

so let the maximum height it reached be h

mgh=1/2mv^2=1/2mgh0

so h=1/20

Is this right? Thanks
 
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  • #2
Yes, looks good. (I presume you mean h = h0/2.)

when goin up the perfect smooth surface it looses the ability to roll
Just to be clear (I'm sure you know this), the hoop loses the ability to roll without slipping. Since there's no friction to exert a torque, the angular speed of the hoop remains constant as it slides up the incline.
 
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  • #3
Doc Al said:
Yes, looks good. (I presume you mean h = h0/2.)

Oops sorry about the typo, yes I mean1/2h0.

Doc Al said:
Since there's no friction to exert a torque, the angular speed of the hoop remains constant as it slides up the incline.

Is this constant angular speed as it slides up the incline equal to the final augular speed it obtained during the rolling down from the other slope?

I don't know if this is right, but is it because rolling and sliding happen at same time, so although it is rolling and we can see it (or can we?) but it doesn't contribute to the motion of the hoop up the incline? SO on a perfect smooth and level ground if we take away the translational motion the hoop will just roll at where it is without going anywhere? Also is that the other half of the initial total energy "lost" due to this sliding motion, the rotational energy is there but it is not contributing to the motion up the slope? Sorry about all those questions:redface: Thanks
 
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  • #4
phalanx123 said:
Is this constant angular speed as it slides up the incline equal to the final augular speed it obtained during the rolling down from the other slope?
Absolutely.

I don't know if this is right, but is it because rolling and sliding happen at same time, so although it is rolling and we can see it (or can we?) but it doesn't contribute to the motion of the hoop up the incline? SO on a perfect smooth and level ground if we take away the translational motion the hoop will just roll at where it is without going anywhere?
Yes. With no friction, the rotational and translational motions are independent.

Also is that the other half of the initial total energy "lost" due to this sliding motion, the rotational energy is there but it is not contributing to the motion up the slope?
When friction is present, and something "rolls without slipping", the rotational and translational motions are coupled--they increase and decrease together. Without friction, there is no coupling. As the hoop rolls up the frictionless ramp, the only force acting on the hoop parallel to the ramp is gravity. Gravity reduces the translational speed, but not the rotational speed, since gravity exerts no torque on the hoop. So the rotational energy remains unchanged.
 

Related to Another Solution check Please. (cylinder rolling down a slope)

1. What is the purpose of the experiment?

The purpose of the experiment is to test the hypothesis that a cylinder will roll down a slope at a constant acceleration, regardless of its mass or shape.

2. What materials are needed for the experiment?

The materials needed for the experiment include a cylindrical object (such as a can or bottle), a slope or ramp, a measuring tape or ruler, a stopwatch, and a flat surface to conduct the experiment on.

3. How should the experiment be set up?

The experiment should be set up by placing the slope or ramp on the flat surface, ensuring that it is stable and at a consistent angle. The cylindrical object should then be placed at the top of the slope and released, while the stopwatch is started. The distance traveled by the cylinder in a certain amount of time can be measured and recorded.

4. What variables should be controlled in the experiment?

The variables that should be controlled in the experiment include the mass and shape of the cylindrical object, the angle of the slope, and any external forces that may affect the cylinder's movement (such as wind or friction).

5. What conclusions can be drawn from the experiment?

From the experiment, we can conclude that the cylinder does indeed roll down the slope at a constant acceleration, regardless of its mass or shape. This supports the theory of inertia and the concept of gravitational acceleration, as the cylinder's acceleration is solely dependent on the angle of the slope and not its mass or shape.

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