Rotational motion and the normal force

[solaris123]

I concluded that ((mv^2)/r)-(mg) was equal to the normal force but the answer says its equal to ((mv^2)/r)-(2mg). Could someone explain why to me please?

"A half sphere of radius r is fixed to a horizontal floor. The inner surface is smooth. A small ball of mass m is at the bottom of the sphere and is given a velocity v. The initial velocity v is sufficiently large so that the small ball reaches point Q at the height r. Find the magnitude of the normal force exerted by the sphere on the ball when the ball passes through point Q. The acceleration due to gravity is equal to g."
Possible answers
(a) (mv^2)/r: not the answer because it doesn't account for gravity
(b) ((mv^2)/r)-(mg): my answer because it accounts for gravity
(c) ((mv^2)/r)-(2mg): the right answer but no clue why
(d)-(h): Other answers which use the kinetic energy formula and thus are not relevant

 

[tiny-tim]

Welcome to PF!

Hi solaris123! Welcome to PF!

(try using the X² tag just above the Reply box )

(b) ((mv^2)/r)-(mg): my answer because it accounts for gravity

You've given no clue as to how you got your formula, ie how it accounts for gravity.

Perhaps you're misunderstanding the question? The ball has a speed w, say (not zero), when it reaches Q.

Anyway, you do need to use the kinetc energy formula.

Try again. ​

 

[solaris123]

No, the answer can't contain the kinetic energy formula because they are asking for the normal force at point Q, having the kinetic energy formula on the answer then makes no sense.
As to how I got my answer, the formula upthere is the centripetal force minus the weight of the object which would have reduce the velocity of the object and thus affect the centripedal force.
I know the magnitude of the normal is equal to the centripedal force because the centripedal force acts toward the center of whatever rotational system, making there be a balancing force ( a la second of Newton) directed outward and because the sphere is there, the sphere counters this force applied by the ball, balancing the ball yet again against the force that arosed from countering the centripedal force, making the normal force of the ball to the sphere at point Q be equal to the centripedal force which is in turn is equal (m*v^2)/r.
However this would only work if we only had a rotational system, but because the ball is working against gravity the velocity at that point Q is less than at point P at the floor. So now we must eliminate part of the value of the centripedal force and because we know the change in height, r, we can use the kinetic energy-potential energy relationship to calculate work and then calculate force and because the change potential energy is r and displacement is r then both cancel and you end up with mg being the part of the velocity you eliminate, leading to answer b rather than answer c.
So now that I have shown I don't just want the answer and did give this a thought could you please answer my question?

 

[tiny-tim]

Hi solaris123!

As to how I got my answer, the formula upthere is the centripetal force minus the weight of the object which would have reduce the velocity of the object and thus affect the centripedal force.

A horizontal force minus a vertical force? How could that be right?

I know the magnitude of the normal is equal to the centripedal force because the centripedal force acts toward the center of whatever rotational system, making there be a balancing force ( a la second of Newton) directed outward and because the sphere is there, the sphere counters this force applied by the ball, balancing the ball yet again against the force that arosed from countering the centripedal force, making the normal force of the ball to the sphere at point Q be equal to the centripedal force which is in turn is equal (m*v^2)/r.

hmm … this is very confused.

The normal force is the centripetal force … it's just another name for it.

"Centripetal" means towards the centre of curvature, which must be in the direction of the normal (or in the opposite direction, of course).

The centripetal acceleration is w²/r (you can't use "v" in this case, because the question has already assigned "v" as the speed at the bottom! )

Using good ol' Newton's second law (F = ma) in the radial direction, we know that a = w²/r, and the only force is N, so N = mw²/r.

So now you have to find w²

btw, the centripetal force is usually not equal to m times the centripetal acceleration … you always have to add (or subtract) the normal component of mg … in this case, it just happens to be zero … so:

WARNING! it is a very bad idea to talk about "centripetal force" …

forget the phrase …

it will (and does) only confuse you

However this would only work if we only had a rotational system, but because the ball is working against gravity the velocity at that point Q is less than at point P at the floor. So now we must eliminate part of the value of the centripedal force and because we know the change in height, r, we can use the kinetic energy-potential energy relationship to calculate work and then calculate force and because the change potential energy is r and displacement is r then both cancel and you end up with mg being the part of the velocity you eliminate, leading to answer b rather than answer c.

ok, I'm not really understanfding this except in general terms.

Yes, as you say, you need the kinetic energy-potential energy relationship (which btw is what I meant by "kinetic energy formula") …

can you show us what that is, and how you applied it to find w²?

 

[D H]

No, the answer can't contain the kinetic energy formula because they are asking for the normal force at point Q, having the kinetic energy formula on the answer then makes no sense.

It makes a lot of sense in this case. Centripetal force is mv²/r.

 

[solaris123]

I found, the change of v is the answer to my conondrum. Applying formula W^2= (V^2)-2ar (the change of speed vertically we get the answer, thank you for your suggestions.