I Why normal force is not equal to gravity?

[user079622]

You conclude that the net force on the mass is zero. You also know that one of the forces on the mass is the tension in the string and that it is directed towards you. What would you say is the other force that balances the tension and points away from you? Where does that force come from?

I would say centrifugal force, from nowhere.

 

[user079622]

@jbriggs444

centrp. acc for moon from revolution (r= moon radius +distanace to earth) = 0.0027 m/s2 and from rotation= 0.0000124 m/s2
Its weird that these result are not the same,because they rotate at same rate.

I dont need to added this two numbers to get total centripetal acceleration on dark side of moon?

Do I need add this two numbers if moon rotate around itself faster, 50rotation/per 1month?

centr. force due to revolution at visible side act away from moon surface and on dark side act toward surface?

 

[jbriggs444]

@jbriggs444

centrp. acc for moon from revolution (r= moon radius +distanace to earth) = 0.0027 m/s2 and from rotation= 0.0000124 m/s2
Its weird that these result are not the same,because they rotate at same rate.

I dont need to added this two numbers to get total centripetal acceleration on dark side of moon?

Do I need add this two numbers if moon rotate around itself faster, 50rotation/per 1month?

centr. force due to revolution at visible side act away from moon surface and on dark side act toward surface?

No. Do not add them. That would be double dipping. If you consider the revolution of the moon about the earth as a rigid body, you have already accounted for all of the motion.

 

[user079622]

No. Do not add them. That would be double dipping. If you consider the revolution of the moon about the earth as a rigid body, you have already accounted for all of the motion.

But if moon rotate very fast,(so no synchronous rotation) we have now two sources of centripetal acceleration, revolution and rotation?
Why not then?

 

[kuruman]

I would say centrifugal force, from nowhere.

Exactly. You asked

why we need introduce fictive forces if object dont accelerate?

and you answered your own question.

In the non-inertial rotating frame the mass at the end of the string is at rest so you say that a "a centrifugal from nowhere" must be acting on it. That force "from nowhere" is fictitious. You made it up because you wanted to satisfy Newton's first law because if the mass is at rest there must be a second force to cancel it. However in so doing, you violated Newton's third law. There is an "action" force on the mass but the mass can exert no "reaction" force on anything because it doesn't exist. Moving mass times acceleration to the left-hand side of Newton's second law does not make it a real force.

 

[user079622]

here is an "action" force on the mass but the mass can exert no "reaction" force on anything because it doesn't exist. .

action force is centripetal?

What you say about my post #34?

 

[kuruman]

action force is centripetal?

What you say about my post #34?

I say that's a different physical situation. I am considering a simpler case to answer your question

why we need introduce fictive forces if object dont accelerate?

and I have answered that.

 

[jbriggs444]

But if moon rotate very fast,(so no synchronous rotation) we have now two sources of centripetal acceleration, revolution and rotation?
Why not then?

You would still be double dipping. The two sources are still not independent.

 

[user079622]

You would still be double dipping. The two sources are still not independent.

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal acceleration as ball that is only spinning 1000rpm?

This sound impossible.

 

[A.T.]

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal force as ball that is only spinning 1000rpm?

This sound impossible.

What do mean by "has same centripetal force"? Maybe draw a diagram and label is properly.

 

[user079622]

What do mean by "has same centripetal force"? Maybe draw a diagram and label is properly.

I mean centripetal acceleration,

Point A and B has same centripetal acceleration?

 

[jbriggs444]

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal acceleration as ball that is only spinning 1000rpm?

This sound impossible.

Be careful. Are you specifying a "sidereal" rotation rate? Or a rotation rate relative to a frame that rotates with the orbit?

Let us assume sidereal. Let us speak of a bug perched on a rigidly spinning ball in a circular orbit about the center of a carousel.

We calculate the centripetal acceleration required to keep the bug on the surface of the spinning ball without accounting for the orbital motion. Simple. That's $\omega^2r$ where r is the ball's radius and $\omega$ is the rotation rate of the ball.

Now we go to add in the orbital motion.

The center of the ball is at one orbital radius. The outer surface of the ball is at a larger orbital radius. The acceleration of the bug relative to the ball due to the orbital motion would seem to be the difference of the two. But wait a minute... If we used this calculation, it would be like assuming that the bug is moving in lock step with the carousel. But it is not. The ball is spinning. The bug is moving with the ball. And we've already accounted for the acceleration associated with that spin.

So to do the calculation correctly, we would need to do the orbital calculation as if the ball has zero sidereal rotation. In that case, the bug's acceleration due to orbital motion would exactly match the acceleration of the center of the ball.

 

[user079622]

The bug is moving with the ball. And we've already accounted for the acceleration associated with that spin.

Tangential velocity and radius of point A and B are different.

why you calculate like there are same?

it could be 2 turn table,frame earth, inertial

small table r=1m, 1000rpm
big table r=10m, 500rpm

point A at small table, ac=10966 m/s2 , tangential velocity=104 m/s

now I am first find what is tangetianl velocity at point B if table is lock to table, so r=11m tan. velocity=576 m/s

I add both tang. velocity 104+576=680 m/s is in point B,when table spin unlock from big table
ac=42036 m/s2

I am not sure if I correct calculate tangential velocity at point B...I strugle in relative motion.
Velocity 104 and 576 are both relative to earth, so I think cant add them,
I turn out that velocity at B due to just "revolution"(lock small table to big one) is faster then velocity of A, I must get velocity at B when small table is spinning

 

[user079622]

@jbriggs444
I post task to Homework help, where it belong.

 

[jbriggs444]

So you want to say that ball that revolving 500rpm and spinning 1000rpm has same centripetal acceleration as ball that is only spinning 1000rpm?

This sound impossible.

What is providing the centripetal acceleration? Gravity or a carousel?

 

[user079622]

What is providing the centripetal acceleration? Gravity or a carousel?

carousel

 

[jbriggs444]

So not relevant to the normal force on the moon.

 

[user079622]

So not relevant to the normal force on the moon.

But I didnt mention normal force for this task.

 

[PeterO]

We learned that normal force equal in magnitude as mg and opposite direction, it is reaction force to gravity.
1.If normal force is not equal gravity, isnt this violate newton 3. law action=reaction?
2. If gravity is higher than normal force , this system has net Force non zero,it means accelerate in dircetion of higher force ,which is not true?
3.Professor say it must be higher otherwise we will not move with earth?
It can be equal, when rotate stone with rope,stone pull your arm with same force as arm pull stone,so not need to be higher..

When an object is placed on a horizontal bench in the class room, we have several pairs of forces, which are reaction forces as mentioned in Newtons 3rd law.
1: The object pushes down on the surface -- the surface pushes up on the object.
2: The Earth pulls down on the object -- the object pulls up on the Earth (effect of gravity)
3: The bench pushes down on the Earth -- the Earth pushes up on the bench.

Each force in those pairs are equal in magnitude and opposite in direction.

If you were to place a 1 kg mass on top of your object, the pair of forces mentioned in #2 do not change, but the upward force on the object MUST increase, as it now supports both the object AND the 1 kg mass

 

[PeterO]

16:23

The presenter made a very misleading statement.
He said the force of gravity had to be slightly more than the normal (reaction) force.
The size of the force of gravity is determined by the mass of the two objects (Earth and Person) and the separation of their centres of mass. We cannot alter any of those, so the force of gravity does not change.
He should have said "the reaction force is slightly less than the force of gravity" (giving the net downward force)
Note: another way to reduce the Normal force, would be to attach a Helium balloon to you - but not one big enough to lift you off the round. (Perhaps a couple of those balloons people take to parties).