[bushindo]

Let's say there's a circular track placed perpendicular to the ground as in the following image

The track has radius r. At the bottom of the track, there is a ball with horizontal velocity v. What is the minimum velocity v such that the ball will travel along the track to make a full circle without falling off the track at any point. Assume that there is no friction, and that the radius of the ball is very small relative to the radius of the track.

[Guest]

a = g(acceleration of gravity) = (v(top)^2)/r

1/2mv(top)^2 + mgh = 1/2mv^2 and h = 2r

v^2 = v(top)^2 + 4gr

v = sqrt(5gr) where g = the acceleration of gravity

Assuming no friction means no rotational energy to account for.

Edited November 24, 2009 by Semper Rideo

[Guest]

minimum velocity at the bottom of the track should be = 2sqrt(rg)

[Guest]

minimum velocity at the bottom of the track should be = 2sqrt(rg)

How did you get that so I can go back and see where I went wrong? And please use spoilers.

[Guest]

Perhaps Jagz didn't include the Kinetic Energy of the ball at the top with the Gravitational Potential Energy?

[Guest]

I think the minimum velocity should be sqrt(rg)

The most critical point is when the ball is at the top of the circle. At this point, the forces acting on the ball are its weight and the centripetal force (not sure if this is exact name of the force... its been a long time)

Weight = mg

Centrifugal force = mv²/r

Since they must both be equal to prevent the ball from falling off the top, v = sqrt(rg)

[plainglazed]

Going with the ole conservation of energy rule and I dont think you can discount the potential energy (mgh or mg2r) at the top of the circle.

[Guest]

Perhaps Jagz didn't include the Kinetic Energy of the ball at the top with the Gravitational Potential Energy?

That must be it.

I think the minimum velocity should be sqrt(rg)

The most critical point is when the ball is at the top of the circle. At this point, the forces acting on the ball are its weight and the centripetal force (not sure if this is exact name of the force... its been a long time)

Weight = mg

Centrifugal force = mv²/r

Since they must both be equal to prevent the ball from falling off the top, v = sqrt(rg)

the v in your equation is the velocity at the top of the circle not at the bottom as asked.

Edited November 25, 2009 by Semper Rideo

[Guest]

I think the minimum velocity should be sqrt(rg)

The most critical point is when the ball is at the top of the circle. At this point, the forces acting on the ball are its weight and the centripetal force (not sure if this is exact name of the force... its been a long time)

Weight = mg

Centrifugal force = mv²/r

Since they must both be equal to prevent the ball from falling off the top, v = sqrt(rg)

Centrifugal force is a pseudo force that doesn't actually exist. To the ball it may seem as though it is being pushed away from the center of the circle by some force when, in reality, an applied force (called a Centripetal Force) is accelerating it towards the center of the circle. It's like when you hit the gas in your car. It feels like you are pushing back on your seat when in fact your seat is pushing forwards on you as you try to remain at a constant velocity. So to find the minimum velocity we want the case where at the top of the circle, gravity is providing 100% of the centripetal force needed to keep the ball moving along the circle defined by the track. Any slower and the ball would fall off the track. Any faster and the tack would need to help a little bit by pushing inwards on the ball. In the end the desired velocity (at least at the top of the circle) for this to be true is indeed v = sqrt(rg), so in this case it seems like an insignificant difference. But it is an important one for more complicated scenarios.

[Guest]

Hi i am curious

a = g(acceleration of gravity) = (v(top)^2)/r

why is it so?

when u use v^2/r you are talking about angular acceleration. It should not be the same as the g. Do correct me if im on the wrong track.

[Guest]

v = 2sqrt(rg)

Technically the velocity must be greater than this but let's not be pedantic.

[Guest]

i think it still needs some momentum at the top so

If the velocity at the top is u then u²/r - g > 0 otherwise it will be in free fall and come off the track.

Therefore u² = rg is the minimum.

With conservation of energy: 0.5*m*v² = mgh + 0.5*m*u²

0.5*v² = 2rg + 0.5*rg

v² = 5rg

v = sqrt(5rg)

[Guest]

Thanks for backing me up Tuckleton and psychic_mind. I think my work might have been too terse. psychic 'splained it better.

Edit: Only at the instant it gets to the top does the angular acceleration need to be equal to gravity's, because an instant later it is being accelerated in a different direction which keeps it on the track.

Edited December 2, 2009 by Semper Rideo