The Tire's Velocity at the Embankment's Top: 2.8 m/s

[huskydc]

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?

 

[OlderDan]

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.

 

[thepatientmental]

It seems like you are on the right track with using energy conservation to solve for the velocity of the tire at the top of the embankment. However, there are a few things to consider in your calculations.

First, make sure that the value you are using for the moment of inertia (I) is correct. The moment of inertia of a solid cylinder (which is what a tire can be approximated as) is given by I = 1/2 * M * R^2, where M is the mass of the object and R is the radius. In your case, I would use I = 1/2 * 10 kg * (0.3 m)^2 = 0.45 kg*m^2.

Secondly, when using the conservation of energy equation, you need to make sure that all the terms have the same units. In this case, your kinetic energy terms have units of Joules (J), while your potential energy term has units of Newton-meters (Nm). To make them consistent, you can convert the potential energy term to Joules by multiplying it by the acceleration due to gravity (9.8 m/s^2). So your final equation should look like this:

[1/2 * 10 kg * V^2 + 1/2 * 0.45 kg*m^2 * (V/0.3 m)^2] - 10 kg * 9.8 m/s^2 * 1 m * sin(30°) = 0

Simplifying this equation, you should get:

0.5 V^2 + 0.075 V^2 - 49 = 0

Solving for V, you should get V = 5.2 m/s. So the tire's velocity at the top of the embankment is approximately 5.2 m/s.

One thing to note is that this calculation assumes that there is no friction or air resistance acting on the tire. If there is friction, it will affect the tire's velocity as it rolls up the embankment. Also, make sure to double check your calculations and units to make sure everything is consistent. Hope this helps!