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#### /0

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##### Circular Motion
« on: April 03, 2008, 10:51:29 pm »
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A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver's mass is 70kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

Thanks.. I can't get this answer

Answer: ${{1.37\times 10^3 \ \mbox{N}}}$
« Last Edit: April 03, 2008, 10:55:32 pm by DivideBy0 »

#### Collin Li

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##### Re: Circular Motion
« Reply #1 on: April 03, 2008, 11:00:52 pm »
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At the top of the hill, the $F_{net} = 700 + 0 = 700\mbox{ N}$ downwards. This is also the force required to keep the person in circular motion. That is, $F_c = F_{net} = 700\mbox{ N}$ down (towards the centre of the circle).

To keep the car moving in circular motion at the bottom of the hill, there must be a $700\mbox{ N}$ force pointing upwards. To do this, $F_{net} = F_c = 700\mbox{ N}$ (where up is defined as positive forces). Since $F_{net} = N - mg$, then:

$\implies N - mg = 700$

Since $mg = 700\mbox{ N}$

$\therefore\;N = 1400\mbox{ N}$

The answer is 1370 because they used 9.8 instead of 10. I will use 10 because VCE Physics uses 10.
« Last Edit: April 03, 2008, 11:02:57 pm by coblin »

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##### Re: Circular Motion
« Reply #2 on: April 04, 2008, 12:17:52 am »
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Cool, so you don't actually need the circular motion formula at all! Thanks coblin

#### enwiabe

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##### Re: Circular Motion
« Reply #3 on: April 04, 2008, 12:20:26 am »
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Coblin: How could it not have a normal force? It's on top of a hill! Hence pushing down on said hill, and hence experiencing a normal force!

#### Collin Li

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##### Re: Circular Motion
« Reply #4 on: April 04, 2008, 12:20:50 am »
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Yeah. The formula for circular motion specifies a force required for a particular speed and radius. In this case, the radius and the speed are the same, so we know that $F_c$ stays constant. We can infer what $F_c$ by using net forces and our knowledge of what it is (the force pointing towards the centre of the circle), and hence we can use it in the second part (the valley) to find what we want!

#### Collin Li

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##### Re: Circular Motion
« Reply #5 on: April 04, 2008, 12:22:01 am »
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Coblin: How could it not have a normal force? It's on top of a hill! Hence pushing down on said hill, and hence experiencing a normal force!

Well the question specified that, so I guess it is instantaneously free-falling. Vertical circular motion is a bit funny, so I wouldn't discount this as impossible.

On another topic, vertical circular motion is not on the course in terms of generalised formulas for it, but something conceptual like this is certainly viable on the mid-year exam.
« Last Edit: April 04, 2008, 12:24:41 am by coblin »

#### enwiabe

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##### Re: Circular Motion
« Reply #6 on: April 04, 2008, 12:24:21 am »
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Or, really, there is a normal force *and* the force required to keep it in circular motion. And it'd actually be, taking downward direction to be positive:
Fc - Nc = 700. And, hence, Fc > 700. But, for the purposes of VCE Physics they just tell them there's no normal force?

#### Mao

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##### Re: Circular Motion
« Reply #7 on: April 04, 2008, 02:47:22 pm »
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is it...?

I thought it was an instantaneous free-fall

when you are travelling on the outside at the topmost point:
imagine the same circular motion without gravity. If the driver is not buckled down to the seat he/she would fly off in a tangent.

with gravity, the person would be accelerating downwards due to it. However, if Fc=Fg (on the person), it implies that the car is accelerating down towards the centre at the same rate. there is no N, since there is no acceleration of the person against the car.

which is also why at the bottom, since the seat is going upwards due to Fc, your acceleration relative to the seat is actually doubled, and N is twice the magnitude.

is that.. right?
« Last Edit: April 04, 2008, 02:50:40 pm by Mao »
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#### Collin Li

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##### Re: Circular Motion
« Reply #8 on: April 04, 2008, 03:03:04 pm »
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A car is driven at constant speed over a circular hill and then into a circular valley with the same radius. At the top of the hill, the normal force on the driver from the car seat is 0. The driver's mass is 70kg. What is the magnitude of the normal force on the driver from the seat when the car passes through the bottom of the valley?

#### Mao

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##### Re: Circular Motion
« Reply #9 on: April 04, 2008, 03:33:40 pm »
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sorry i didnt state my intentions clear

i was commenting how what its like when Fc=mg, (N=0 at top)

trying to clear out this question that left me quite confused...
Fc - Nc = 700. And, hence, Fc > 700. But, for the purposes of VCE Physics they just tell them there's no normal force?
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#### /0

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##### Re: Circular Motion
« Reply #10 on: April 04, 2008, 04:29:06 pm »
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Thanks for help on that problem, I have another one... it's more a concept than an actual question

A puck slides of mass m slides around a frictionless table in a circle, and a string is attached through a hole in the circle's centre to a cylinder of mass M. At what speed must the puck travel so the system is in equilibrium?

What I don't get is this: The puck, due to centripedal force, moves towards the centre, and the cylinder, because of gravitational force, moves down from the centre. Ergo, the net movement must be down the hole. I guessed that the force acting upwards is $\frac{mv^2}{r}$ but I why is this? Shouldn't this force act 'downwards'?
Thankyou again

#### Mao

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##### Re: Circular Motion
« Reply #11 on: April 04, 2008, 04:32:11 pm »
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the puck has a force acting towards the centre (but it never reaches it)

the tension force of the string attached to the puck then must balance the puck's centripedal force.

when this tension force is the same magnitude as the gravitational force on the cylinder, you'll have equilibrium =D

remember that the centripedal force is acting on the puck ONLY.
that is, it has little effect on the cylinder, other than that the opposing tension force of the string can keep it in the air.

in this equilibrium, if the centripetal force Fc is the same as weight of M Fg:
at the puck: Fc is balanced by tension Ft going outwards (or up)
at the cylinder: Fg is balanced by the same tension going upwards.
« Last Edit: April 04, 2008, 04:46:05 pm by Mao »
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##### Re: Circular Motion
« Reply #12 on: April 04, 2008, 06:11:36 pm »
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Thankyou Mao, I haven't revised string tension for a while