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The vector of centripetal acceleration at a point has always the direction of the radius at that point. So in your scheme, the vector of centripetal acceleration points towards the center, has the same direction as ##\vec{F_N}##.
Uniform Circular Motion and Centripetal Acceleration (non-intuitive)
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babaliaris
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Delta² said:
Yes but i don't know Fn I want to calculate it. Please also consider #30 (because i provided you with the wrong image on #27)
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yes I war referring to the second scheme in post #30. The centripetal acceleration there has the same direction as ##\vec{F_N}##. But the total acceleration which is ##\vec{a}=\vec{a}_{tangential}+\vec{a}_{centripetal}## makes an angle with the x-axis which you can calculate, together with the calculation of ##F_N##
babaliaris
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I can't understandDelta² said:
This is where i reached so far:
I know v, m, r and g but not theta. I need theta to calculate the final result...
[Edited]
LoL actually i can find the magnitude of Fn when the particle is upside down and then in the formula above a can replace Fn = |Fn| (cos8 i + sin8 j) and now i have to find (cos8 i + sin8 j) . This is what you meant
(8 = theta symbol)Attachments
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First of all ##\Theta## can be calculated if you know the height h of that point, it will be ##\sin\Theta=\frac{h}{R}##
Your equation is correct at start but at some point it goes wrong. The total acceleration is not only the centripetal acceleration. There is tangential acceleration too. You correctly expanded centripetal acceleration to its i, j components, but you need to add the tangential acceleration components too. So you missing two terms. You also forgot to write g as ##g\vec{j}##.
Your equation is correct at start but at some point it goes wrong. The total acceleration is not only the centripetal acceleration. There is tangential acceleration too. You correctly expanded centripetal acceleration to its i, j components, but you need to add the tangential acceleration components too. So you missing two terms. You also forgot to write g as ##g\vec{j}##.
babaliaris
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Never heard of tangential acceleration yet. I just know that a particle that is doing uniform circular motion has only one acceleration, the centripetal acceleration. This is what the book told me so far. Next chapter is kinetic energy.
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Very well, but in the situation you examining at post #30, the particle does non-uniform circular motion... that's the whole "catch"... So better to avoid this problem till you learn about tangential acceleration.babaliaris said:
babaliaris
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I see... I will wait then until i reach the appropriate chapter. Once again I'm rushing
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+1
It's not often you need to calculate the net acceleration when the motion is non-uniform. Try easier problems first.
It's not often you need to calculate the net acceleration when the motion is non-uniform. Try easier problems first.
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Again, math is your friend. If I understand the problem right, you have a circular loop in the ##x_1 x_2## plane. For convenience I put its center in the origin of the coordinate system. The gravitational acceleration is ##\vec{g}=-g \vec{e}_y##. Now the motion on the circular loop most simply is described with polar coordinates,
$$\vec{r}=r \cos \phi \vec{e}_1 + r \sin \phi \vec{e}_2.$$
The equation of motion is
$$m \ddot{\vec{r}}=m \vec{g} \, \Rightarrow \, \ddot{\vec{r}}=\vec{g}.$$
The only thing you need is the 2nd derivative wrt. time.
Then fortunately the force is conservative, i.e., it has a potential
$$V(\vec{r})=-m \vec{g} \cdot \vec{r}=m g x_2=m g r \sin \phi.$$
Now energy is conserved, i.e.,
$$E=\frac{m}{2} \dot{\vec{r}}^2+m g r \sin \phi=\text{const}.$$
With the chain rule (and because ##r=\text{const}##) you have
$$\dot{\vec{r}}=-r \dot{\phi} \sin \phi \vec{e}_1 + r \dot{\phi} \cos \phi \vec{e}_2$$
and
$$\dot{\vec{r}}^2=r^2 \dot{\phi}^2.$$
Now it should be very easy to solve your problem, using the energy conservation law above.
$$\vec{r}=r \cos \phi \vec{e}_1 + r \sin \phi \vec{e}_2.$$
The equation of motion is
$$m \ddot{\vec{r}}=m \vec{g} \, \Rightarrow \, \ddot{\vec{r}}=\vec{g}.$$
The only thing you need is the 2nd derivative wrt. time.
Then fortunately the force is conservative, i.e., it has a potential
$$V(\vec{r})=-m \vec{g} \cdot \vec{r}=m g x_2=m g r \sin \phi.$$
Now energy is conserved, i.e.,
$$E=\frac{m}{2} \dot{\vec{r}}^2+m g r \sin \phi=\text{const}.$$
With the chain rule (and because ##r=\text{const}##) you have
$$\dot{\vec{r}}=-r \dot{\phi} \sin \phi \vec{e}_1 + r \dot{\phi} \cos \phi \vec{e}_2$$
and
$$\dot{\vec{r}}^2=r^2 \dot{\phi}^2.$$
Now it should be very easy to solve your problem, using the energy conservation law above.
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babaliaris said:
Have you heard of Wikipedia?
https://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration
babaliaris
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Yes but anyway, it's very early for me right now. I just started with physics and I was rushing.A.T. said:
Friend of Kalina
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1. The ball on the string doesn't fall to the center because it starts with a velocity tangent to the circle. The string pulls it into the center of the circle, the velocity imparts an inertial component to keep going straight, the result is that the ball turns along the circumference of the circle.
2. The ball doesn't fall to the ground because the string is holding it up. If you stand to the side, you will see that the string is not parallel to the ground, but is instead at an angle. The total force on the string is composed of the horizontal acceleration, keeping the ball going in a circle, and the vertical acceleration countering gravity. The flatter you keep the string, the higher the force required; centripetal force is linked by the angle of the string, but also increases, and the ball goes faster.
Oops. Somehow failed to notice I was still on page 1 of 3 and that question had been resolved. Sorry.
2. The ball doesn't fall to the ground because the string is holding it up. If you stand to the side, you will see that the string is not parallel to the ground, but is instead at an angle. The total force on the string is composed of the horizontal acceleration, keeping the ball going in a circle, and the vertical acceleration countering gravity. The flatter you keep the string, the higher the force required; centripetal force is linked by the angle of the string, but also increases, and the ball goes faster.
Oops. Somehow failed to notice I was still on page 1 of 3 and that question had been resolved. Sorry.
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You probably have the answer but...babaliaris said:
When a car accelerates forward, you feel like you are being pushed, back into the seat.
When a car brakes heavily in an emergency stop (it is accelerating back) you feel you are being thrown forward,
Why should it feel unusual that when a truning car is accelerating towards the centre of the circle (centripetal acceleration) you feel like you are being thrown outwards, away from the centre.
The real truth is that you are not being pushed or thrown in any of those directions - it just feels like you are. We are often deceived. We can even feel (momentarily) lighter when an elevator starts to move, then (momentarily) heavier when it stops again.
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No! What you feel in the car is that you get pushed forward by the force imposed on you by the seat!PeterO said:
It's the same thing in a merry-go-around: You feel the push of the force imposed on you by the seat towards the center of the circle, i.e., the centripetal force.
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"Feel" can mean different things:vanhees71 said:
- Direct sensation of contact forces (what you probably mean)
- Interpretation of your movement relative to the car (what Peter probably means)
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Maybe a physicist "feels" differently from being trained to think differently. ;-)).
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In a carnival scrambler ride, there's a lot of centripetal acceleration, but 2 people sit side by side in the ride with no seat belts. If the riders end up pushing against each other, the outside rider exerts a centripetal force on the inside rider, and the inside rider exerts a centrifugal reaction force to the outside rider. This is one example where a centrifugal reaction force can be felt.
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I am quite aware of what forces are actually acting on you - I was using "feel" as the sensation you experience. Indeed if you watch an unbelted child sitting in the back seat of the car they do not feel a centripetal force from the seat - they eventually feel the centripetal force from the door when the have finally tumbled that far. They don't know to hang on - especially if they have been brought up to always wear a seat belt and have just omitted this time. The seat belt can supply the centripetal force need without the "outward tumble" - which is, I know, just continuing (approximately) in a straight line.vanhees71 said:
By the way, another way to make the seat push forward on you is to have someone else push you into the seat - while the car is stationary. It that case you still don't notice the seat pushing, but wonder why the person is pushing you. When the car accelerates, you still don't notice the seat pushing, you just wonder who is pushing you into the seat this time. One reason to study Physics is so you can explain that there is nothing pushing you into the seat, it is just a ficticious force that has appeared in the non-inertial frame you have briefly found yourself in.
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When a car is accelerating, you have a Newton 3rd law pair of forces, the seat exerts a forwards force on the person, the person exerts a reaction force on the seat. The reaction force is a real force in any frame of reference.PeterO said:
Using the car as an accelerating frame of reference, the fictitious force is the force that would appear to accelerate a body in "free fall" with respect to the accelerating frame of reference, and would not be part of a Newton 3rd law pair of forces. The Newton 3rd law pair of forces are still the same, seat exerts forward force on person, person exerts backwards force on seat.
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Yeah, we must not forget how the seat feels about it.rcgldr said:
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Nothing wrong with those Newtons 3rd law Forces, other than why they are happening.rcgldr said:
I contend a person only expects to apply a force on the seat, if they decide to push, or can simply explain why they are doing it.
In a stationary car, If a third person pushes an occupant, the occupant is not surprised that they therefore apply/transfer a push to the seat, and they will feel the seat pushing back. That person does NOT feel themselves pushing back on the third person - only the third person feels that force.
When the occupant is in the accelerating car, they feel the seat pushing forward, but may be at a loss to explain why the seat is doing that - because in that situation there is no third person instigating the action/reaction. Unfortunately, after just a few seconds, the car ceases to accelerate and the occupant quickly loses interest in the situation except to think "wow, something really pushed me into that seat just then", and perhaps include "it must have been the engine". Remember, we are not talking about a physics person explaining what is happening, we are talking about an "innocent" wondering what just happened.
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A person undergoing acceleration is going to feel real forces internal to their body and feel compression forces at the points of contact with the seat. This is more clear in the case of greater acceleration such as a graviton / rotor spinning carnival ride that pins the persons (victims) against the wall while the floor drops, or in a vehicle capable of pulling more than 1 g.PeterO said:
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Yes they will feel the force, but if they are a non-physics person (as are our students when they begin to stud the subject) they won't necessarily understand why that force suddenly came about. Seats and walls don't usually start pushing you unless you start pushing them - in these cases, the seat/wall seems to instigate the entire action/re-action.rcgldr said:
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I think most non-physics persons get a sense of these forces when riding roller coasters that include high g (accelerating) turns.PeterO said: