Is F = m*(v^2/r) true when the orbital speed increases?

[Karagoz]

In circular motions, one can measure the NetForce on an object with this formula:

ΣF = m*v^2/r.

But is this formula valid even if the orbital speed of the object is constantly increasing (or constantly decreasing)?

 

[BvU]

Yes. v is the instantaneous speed
See #6 and #7

 

[ZapperZ]

In circular motions, one can measure the NetForce on an object with this formula:

ΣF = m*v^2/r.

But is this formula valid even if the orbital speed of the object is constantly increasing (or constantly decreasing)?

This requires A LOT more explanation of the scenario.

Is this increase in speed of an orbiting object where the trajectory isn't fixed? Because if it is, then an increase in speed will also cause a change in the radius, thus, there will be a radial component to the velocity during this change.

If this is strictly at a fixed radius (a centrifuge spinning faster and faster), then yes, that formula basically still works at a particular instant.

Zz.

 

[jbriggs444]

If this is strictly at a fixed radius (a centrifuge spinning faster and faster), then yes, that formula basically still works at a particular instant.

With the caveat that it is the instantaneous radial component of the acceleration that is calculated. Any tangential component required to achieve the increasing or decreasing speed will be in addition to the radial component.

 

[A.T.]

In circular motions, one can measure the NetForce on an object with this formula:

ΣF = m*v^2/r.

But is this formula valid even if the orbital speed of the object is constantly increasing (or constantly decreasing)?

This is the formula for the radial component. It equals the net force only if there is no tangential component. Is that the case in your scenario?

 

[Karagoz]

This is the formula for the radial component. It equals the net force only if there is no tangential component. Is that the case in your scenario?

In my scenario, the radius and the form of the "circle" remains the same. But the orbital speed of the object do increase constantly. So there's a tangential acceleration, hence a tangential force.

So the "centripetal force" doesn't point towards the center of the circle, as in the picture below (red arrow show ΣF, NetForce).

The formula ΣF = m*v^2/r is it valid if the orbital speed of the object do increase?

 

[BvU]

Should have provided this context in the very first posting !

This case: No, mv²/r is only the component of the force in the radial direction.

 

[Karagoz]

Should have provided this context in the very first posting !

This case: No, mv²/r is only the component of the force in the radial direction.

In a vertical circular motion, the gravity force will pull the object all the time downwards. So the orbital velocity and acceleration does change all the time.
But the minimum orbital velocity the object needs to have to still be in a vertical circular motion is calculated that way:

which gives

(Taken from a physics book, m is for mass, g for gravitational acceleration (ca. 9.81m/s^2), v for velocity, r for radius of the circle).

So even when the "orbital velocity" does change in vertical circular motion, the formula "m*v^2/r" is used to find the required minimum velocity at the top.

Like in the picture below, a pendulum in vertical circular motion.

The net-force won't point towards the center of the "circle" when the object is in that position.

But the net-force still can be calculated with "ΣF = m*v^2/r" even when the object is in that position?

 

[BvU]

Obviously: no !

If, for example, v = 0 at the position in the drawing, the net force is mg and not 0.

 

[Herman Trivilino]

In circular motions, one can measure the NetForce on an object with this formula:

ΣF = m*v^2/r.

That's the formula for the net force exerted on a particle undergoing uniform circular motion.

But is this formula valid even if the orbital speed of the object is constantly increasing (or constantly decreasing)?

If the speed is changing then there is also a tangential component to the net force. The above formula gives you the radial component only.