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Yes. The object's path will curve more sharply than the track, so its path and the track will diverge, i.e., it falls off.
I think you understand this, but just to reiterate BruceW's point, the normal force would never actually cause the roller coaster to come off the track. It would just be enough to keep roller coaster following the circular path. It's just like the normal force exerted on a book by a table is just enough to keep the book from falling downward. It would never be so large as to cause the book to accelerate upward and jump off the table. In other words, your hypothetical situation that mg+reaction > mv2/r (with a non-zero reaction force) is physically impossible.jsmith613 said:ok so if 20 kg moves at a speed of 10 m/s and a radius of 3m then the NET force i need (Fc) is 666.666N
So normally the reaction force should accommodate this? right? but then if moving on the inside of the circle we can say that mg + reaction force = mv^2/r so the mv^2/r would only be the same if the reaction force provides all the rest of the force.
what if the mg + reaction force > 666.666N what would happen then?
Doc Al said:OK. To maintain contact on the inside of a vertical loop, you need a minimum speed of sqrt(gr) at the top; that translates to a minimum speed of sqrt(5gr) at the bottom to make it to the the top with sufficient speed.
mv^2/r is the net force towards the center on something that is executing circular motion. The forces acting on the body will be gravity and the normal force.
Right. There's no way to travel on the outside bottom of a loop without something to pull you up. Normal force and gravity won't help you.jsmith613 said:I know this discussion was along time ago but I want to check something
If I am moving around the OUTSIDE OF a circle, surely at the bottom of the circle (UNLESS I have a second pair of wheels to give me a reaction force big enough to stay on the track) I HAVE TO FALL OFF (regardless of my speed)
Is this right?
Doc Al said:Right. There's no way to travel on the outside bottom of a loop without something to pull you up. Normal force and gravity won't help you.
Right.jsmith613 said:If the object moved at 2 m/s then the centripetal force required is 0.4N
Which is way too much!the force provided at the top is 10N
Don't think in terms of minimum centripetal force, but minimum normal (reaction) force. If the reaction force goes below zero, that means you are losing contact. (And the reaction force can only be positive.)surely this means the object should STAY in the circle as the minimum force is provided. But this seems to contradict the rule min speed is sqrt(gr)
where have I gone wrong??
Doc Al said:Right.
Which is way too much!
Don't think in terms of minimum centripetal force, but minimum normal (reaction) force. If the reaction force goes below zero, that means you are losing contact. (And the reaction force can only be positive.)
The minimum speed is what you need to give you just exactly zero reaction force at the top; no problem if you go faster, but you'll fall off the loop if you go slower.
Doc Al said:Yes.
Another way to understand what's going on when you're inside the circle at the top is to realize that the minimum force on you will equal your weight. So any speed which requires less than that amount of centripetal force cannot keep you on the track. The excessive force of your weight will pull you off the track and into the air.
jsmith613 said:The most terrible thing about this topic is every time I revise it or revist this discussion I always get confused!
With the exam coming up I am now revisiting this topic.
Object moving around the inside of the track
So we discussed earlier that mv^2/r (centripetal force) is the net force towards the center on something that is executing circular motion
This force needs to be provided by associated forces such as reaction force or weight
At the top of the circle, if the radius is 10m and the speed is 30 m/s then the centripetal force required to keep the object (of mass 1kg) in a circle is 90N
the weight of the object is 10N so I would presume the reaction force provides the other 80N.
If the object moved at 2 m/s then the centripetal force required is 0.4N
the force provided at the top is 10N
surely this means the object should STAY in the circle as the minimum force is provided. But this seems to contradict the rule min speed is sqrt(gr)
where have I gone wrong??
thanks
PeterO said:So near and yet so far!
eg: a lift accelerates up at 2m/s/s - how heavy does the 100kg occupant feel? [use g=10 for simplicity]
Net Force [red arrow] 200N UP
Weight Force [black arrow] 1000N DOWN.
Green arrow is thus 1200N up
1200N up is how strongly a floor normally pushes on a 120kg person, so the occupant's apparent weight is 120kg.
Indeed the colour coding is useful in many problems: get your self three different pens and use them always.
Often people forget the weight force in a problem, but given that the first thing you always draw is the black arrow, there can be no omissions.
jsmith613 said:Surely green = 800N
Weight = Acceleration + Reaction force
R = W - A
R = 1000 - 200
R = 800N
so the reaction force will be 800N
PeterO said:NO!
When a lift accelerates up - ie it first moves off when you are traveling up - do you feel heavier or lighter?
The weight force is down - black arrow
The net force is up - the Red arrow
The Green arrow goes from the bottom of the Black arrow, all the way up to the top of the black arrow, then even further to the top of the Red arrow.
As for "Weight = Acceleration + Reaction force" I was simple using
Apparent Weight = Reaction Force
Perhaps your "Weight - acceleration" should be accounting for the fact that acceleration is in one direction, while gravity is in the other, so one of them should be negative
So 1000 - -200 = 1200 ?
When I just use the numbers, I sometimes go wrong.
When I use the Arrows, and look at what is going where, I always get the right answer.
The net force is always in the direction of the acceleration, not necessarily in the direction of the velocity. (In this example, the acceleration and velocity of the lift are both up.)jsmith613 said:why is the net force up? the net force is not always in the direction of motion??