# Centripetal acceleration problem (box sitting on a table)

• Physics guy
In summary: But for stationary masses, the effective g stays the same because the gravitational force is always in proportion to the mass.
Physics guy
Let me imagine a box placed on a table. It has got no acceleration. If I were a person who trusted Newton's laws then I would argue that the net force on the box should be zero. Now in another situation I am an observer outside the Earth and I see that the box is rotating along with the earth, so it should have a centripetal acceleration and gravity provides it...but in the previous case, gravity was canceled by the normal reaction. So shouldn't the centripetal force also be zero. Please explain where I went wrong.

You've got the right intuition, but let's iron a few things out.

For a minute, forget about the Earth. Just imagine an idealised flat, rigid surface of infinite extent. Impose a uniform gravitational field ##-g \hat{\mathbf{y}}##. The book sitting on the table has its weight exactly canceled by the normal force exerted by the table, to satisfy the equilibrium condition.

If the Earth wasn't rotating, that would be a pretty good local model. In fact, even though the Earth is rotating, the rate of rotation is sufficiently slow that this is an approximately correct model.

But we might want to think about a different, maybe more accurate, model in which the Earth has non-zero rotational speed. To simplify the description, imagine the table is at the equator. As you correctly deduced, the net force on the book must now be non-zero and pointing toward's the Earth's centre, in order to provide the necessary acceleration ##\rho \omega^2## toward the centre. In other words, the magnitude of the weight will slightly exceed the magnitude of the normal force.

These two descriptions aren't in contradiction, simply because they're not describing the same scenario. The assumptions that underly the two models are different.

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vanhees71
Physics guy said:
Let me imagine a box placed on a table. It has got no acceleration. If I were a person who trusted Newton's laws then I would argue that the net force on the box should be zero. Now in another situation I am an observer outside the Earth and I see that the box is rotating along with the earth, so it should have a centripetal acceleration and gravity provides it...but in the previous case, gravity was canceled by the normal reaction. So shouldn't the centripetal force also be zero. Please explain where I went wrong.
Newton's Laws apply to inertial frames. The rotating rest frame of the Earth is not exactly inertial. To make Newton's 2nd Law applicable there, you have to introduce inertial forces (here Centrifugal force):

Rotating rest frame of the Earth:
Gravity + Centrifugal + Normal = 0

Inertial frame where the Earth rotates:
Gravity + Normal = Centripetal

https://en.wikipedia.org/wiki/Rotating_reference_frame

Physics guy and vanhees71
To go along with what @A.T. said, in the non-inertial local frame you have the fictitious centrifugal force and the gravitational force. Since they are both proportional to mass you can simply add them together to get an overall “effective” g. This effective g will change from place to place over the globe.

jbriggs444 and vanhees71

## 1. What is centripetal acceleration?

Centripetal acceleration is the acceleration experienced by an object moving in a circular path. It is always directed towards the center of the circle and its magnitude is equal to the square of the object's speed divided by the radius of the circle.

## 2. How does centripetal acceleration affect a box sitting on a table?

When a box is sitting on a table, it is not moving in a circular path. Therefore, there is no centripetal acceleration acting on the box. The only acceleration acting on the box is the force of gravity, which is directed towards the center of the Earth.

## 3. Can centripetal acceleration cause a box to fall off a table?

No, centripetal acceleration alone cannot cause a box to fall off a table. In order for an object to move in a circular path, there must be a force acting towards the center of the circle. If the box is sitting on a table, the force of gravity is acting downwards and the normal force from the table is acting upwards, keeping the box in place.

## 4. How is centripetal acceleration calculated for a box on a table?

Since there is no centripetal acceleration acting on a box sitting on a table, there is no need to calculate it. However, if the box were to start moving in a circular path, the centripetal acceleration can be calculated using the formula a = v^2/r, where a is the centripetal acceleration, v is the speed of the object, and r is the radius of the circle.

## 5. Can the mass of the box affect the centripetal acceleration?

Yes, the mass of the box can affect the centripetal acceleration if the box is moving in a circular path. According to the formula a = v^2/r, the centripetal acceleration is directly proportional to the square of the speed of the object. Therefore, a heavier box with a larger mass will require a greater centripetal acceleration to maintain the same speed in a circular path compared to a lighter box with a smaller mass.

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