Rolling uphill

  • Thread starter huskydc
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  • #1
huskydc
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a tire, rolls up on a ramp,

with a speed of 2.8 m/s, a tire mass of 10 kg and tire radius of 30 cm

the tire cruises up an embankment at 30° for 1 meter

What is the velocity of the tire at the topof the embankment in m/s?


I tried the following:

energy conservation: and end up with something like this:

Final KE =

[ 1/2*M*V^2 + 1/2*I*(V/R)^2 ] - M*g*h (where h = 1m * sin30° )

and I = .45

but I'm not sure if at Final KE, that if there is still the rotational energy, however, I tried solving for V

with both either KE as just .5 m v^2 and .5mv^2 + .5 I (v/r)^2


neither works...hints?
 

Answers and Replies

  • #2
OlderDan
Science Advisor
Homework Helper
3,021
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huskydc said:
a tire, rolls up on a ramp,

with a speed of 2.8 m/s, a tire mass of 10 kg and tire radius of 30 cm

the tire cruises up an embankment at 30° for 1 meter

What is the velocity of the tire at the topof the embankment in m/s?


I tried the following:

energy conservation: and end up with something like this:

Final KE =

[ 1/2*M*V^2 + 1/2*I*(V/R)^2 ] - M*g*h (where h = 1m * sin30° )

and I = .45

but I'm not sure if at Final KE, that if there is still the rotational energy, however, I tried solving for V

with both either KE as just .5 m v^2 and .5mv^2 + .5 I (v/r)^2


neither works...hints?
It appears you are on the right track. There is initial kinetic energy of translation and rotation, and final kinetic energy with the difference between them being the potential energy. The kinetic energy will always be a combination of both translation and rotation as long as the tire is moving.

It appears you are using the equation for a disk to find I. A tire is not a disk. A reasonably accurate model would be a hoop or ring, with all the mass at approximtely the same radius. That is not precise, but it may be the assumption you are expected to make.
 

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