- #26

- 182

- 1

so which is the final answer?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter physior
- Start date

- #26

- 182

- 1

so which is the final answer?

- #27

- 92

- 4

We can't/won't just give you the answer. We'll help you get to it yourself.

- #28

- 182

- 1

this is what you said

but what about the distance covered? is this less too?

- #29

- 92

- 4

The velocity with which the cylinder will move down the hill is ##\omega \times \text{radius}. ##The formula for rotational energy is analogous -- it is ##\frac{1}{2}I\omega^2## where ##I## is the "moment of inertia" of the object and ##\omega## is its rotation rate, also known as "angular velocity". The angular velocity is normally measured in radians per second (this makes the units come out nice). For a given shape, the moment of inertia is proportional to its mass and is proportional to the square of its size.

If you work out the proportions, a cylinder of a given material rolling at a fixed velocity whose radius is doubled...

1. Rotates half as fast due to the increased radius and has 1/4 of the kinetic energy due to that reduced rotation rate.

2. Has four times the moment of inertia due to the increased radius and 4 times the kinetic energy due to that increased moment of inertia.

3. Has four times the mass due to the increased radius and has 4 times the kinetic energy due to that increase mass.

The net increase in rotational kinetic energy is a factor of four. That exactly matches the increase in mass. Just like the increase in linear kinetic energy exactly matches the increase in mass. The two remain proportional to one another!

[Edit, first draft of this missed the increase in moment of inertia]

Distance traveled is ## v \times \Delta t## so the distance is ##\omega \times r \times \Delta t##.

Does that help, along with the post I quoted from someone else above?

Or are you trying to solve this via energy conservation?

- #30

- 182

- 1

so, from what I see, there isn't a single equation to help us answer this question?

- #31

- 92

- 4

- #32

- 92

- 4

I just read this and your answer. Ignore my other post. I think energy conservation would be a lot easier to solve the problem with. But I could be wrong. I suggested the ##\Delta t## after reading the problem statement as that would be the time it takes for the cylinder to reach the bottom of the ramp from Top. Are you confused about how to use the math, physics, or the process, or all of the above?Do you mean there is static contact friction that makes it roll, but no rolling resistance that dissipates energy? If that's the case, then you can use energy conservation to find the speeds.

- #33

A.T.

Science Advisor

- 11,290

- 2,678

Yes, and I think jbriggs444 already did that in post #25.I think energy conservation would be a lot easier to solve the problem with.

Share: