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VCE Stuff => VCE Science => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Physics => Topic started by: stonecold on January 23, 2010, 09:22:56 pm

Title: random physics questions.....
Post by: stonecold on January 23, 2010, 09:22:56 pm: EDIT: If you have any physics questions and don't want to make a new thread, ask them here :)

hey all, trying to understand physics but atm, it seems to be a lost cause.

can someone help me out with this:

a toy car of mass 0.40kg travels down a smooth track banked down at an angle. there is a vertical circular loop of radius 0.24m at the bottom of the track.

a) what is the minimum speed the car must have at the top of the circle to remain on the track?
b) what is the corresponding speed at the bottom of the track for this to occur?
Title: Re: stonecold physics thread.....
Post by: dejan91 on January 23, 2010, 09:50:13 pm: Ok, I'm REALLY rusty on this stuff so here goes nothing:

In the inside of a loop F_sum = N + mg (where down is taken as positive). The minimum speed occurs when N = 0. Thus, F_sum = 4 N

Now, $F_{sum}=\frac{mv^{2}}{r}$
$v = \sqrt{\frac{F*r}{m}}$ where F = 4N, r = 0.24kg, and m = 0.40kg
So $v = 1.55 ms^{-1}$
Title: Re: stonecold physics thread.....
Post by: stonecold on January 23, 2010, 09:52:40 pm: i knew it had something to do with that formula, but how did you get the force to be 4?

....and the mass to be 0.24kg.

sorry, i suck at this!

edit. dont worry, i finally got it :)
Title: Re: stonecold physics thread.....
Post by: dejan91 on January 23, 2010, 10:01:30 pm: Ok, well as I said, in the inside of a loop I take $\sum F = mg+N$ and down to be the positive direction. You know that $mg = 0.4 * 10 = 4N$ . Now, for your speed to be a minimum, you must be at a point where you have little to no contact with the surface (just on the verge of falling down). Of course, this is the reaction force, N. So $N = 0$ and thus $\sum F = 0 + 4 = 4N$ . If this doesn't help, try to visualise a roller coaster going in a loop: if it goes fast enough, you can imagine it pushing the rails upwards. If it were to go too slow, it would just reach the top and fall down. You're finding the speed at which it's about to fall - somewhere in between.

Btw, sorry that m = 0.24kg was a typo.
Title: Re: stonecold physics thread.....
Post by: dejan91 on January 23, 2010, 10:11:24 pm: Next one: use the conservation of energy.

At the top of the loop:
$U_g = mgh = 1.92J$
$E_k = = 0.48J$ , where $v=\sqrt{2.4}$

So, $\sum E = 2.4J$

At the bottom, let $E_k = 2.4J$
$E_k = \frac{1}{2}mv^2$
$\therefore v=\sqrt{\frac{2E_k}{m}} = 3.46ms^{-1}$
Title: Re: stonecold physics thread.....
Post by: stonecold on January 23, 2010, 10:21:18 pm: thank so much. i should have recognized that, because i have done kinetic and gravitational potention energy before.
my textbook doesn't cover it until the next chapter, so i plain forgot about it lol.
Title: Re: stonecold physics thread.....
Post by: QuantumJG on January 23, 2010, 10:41:34 pm: Firstly let's try and visualise what forces allow this car to move in a circular motion along the verticle loop.

Now the two forces in this case for the net force or centripetal force is the weight force and the normal force.

i) At the top we have weight down and the normal force down, so:

$m \dfrac{v^{2}}{r} = mg + n$

Now the minimal speed at the top can be figured out by letting n = 0. Why? Because at the minimal speed you can imagine the car just about to fall off and since the normal force represents how attached the car is to the surface, then at the minimal speed it's 0.

$\rightarrow \dfrac{v^{2}}{r} = g$

$\rightarrow v = \sqrt{ rg }$

$\therefore v_{top min} = 1.53m/s$

ii) Now for the bottom and I'm going to try energy conservation.

$\dfrac{1}{2} m v_{top min}^{2} + m g h_{top} = \dfrac{1}{2} m v_{bottom min}^{2} + m g h_{bottom}$

$\rightarrow v_{top min}^{2} + 2 g h_{top} = v_{bottom min}^{2} + 2 g h_{bottom}$

let h_bottom = 0, h_top = 0.48m, (v_{top min})² = 2.352 m²s^-2

$\rightarrow 11.76 = v_{bottom min}^{2}$

$\rightarrow v_{bottom min} = 3.43m/s$

Hope this helps.
Title: Re: stonecold physics thread.....
Post by: stonecold on January 23, 2010, 10:56:16 pm: thanks guys.

i have another question sorry.

A girl is swinging around a pole in a playground. She is attached to the top of the pole by a rope. The girl has mass 36kg, and the rope makes an angle of 20 degrees with the vertical (the pole). The girl moves in a circlular motion around the pole, with radius 'r'.

a)What is the verticle component of the tension in the rope
b)What is the horizontal component of the the tension in the rope
c)What is the tension in the rope
d)What is the net force acting on the girl
e)What is the value of 'r'.

Also, because i really seem to be struggling, do you think i should maybe just drop physics and focus on my other subjects.
I don't really know what has happened. I did really well with year 11 physics, but this year 12 stuff is really testing me out. i am quite lost.

thanks.

edit: part a should be enough btw, i might be able to do the rest after i see how that is done.
Title: Re: stonecold physics thread.....
Post by: QuantumJG on January 24, 2010, 01:22:20 pm: Quote from: stonecold on January 23, 2010, 10:56:16 pm
thanks guys.

i have another question sorry.

A girl is swinging around a pole in a playground. She is attached to the top of the pole by a rope. The girl has mass 36kg, and the rope makes an angle of 20 degrees with the vertical (the pole). The girl moves in a circlular motion around the pole, with radius 'r'.

a)What is the verticle component of the tension in the rope
b)What is the horizontal component of the the tension in the rope
c)What is the tension in the rope
d)What is the net force acting on the girl
e)What is the value of 'r'.

Also, because i really seem to be struggling, do you think i should maybe just drop physics and focus on my other subjects.
I don't really know what has happened. I did really well with year 11 physics, but this year 12 stuff is really testing me out. i am quite lost.

thanks.

edit: part a should be enough btw, i might be able to do the rest after i see how that is done.

Because the only force that is acting on her is the tension of the rope (I.e. The tension is the centripetal force).

a) if you think about it, the vertical component of the tension must only be able to counteract the weight force.

I.e. T_vert = Tcos(20^o) = mg = 352.8N

b) T_hor = Tsin(20^o) = $\dfrac {352.8 sin(20^{o}) }{ cos(20^{o}) }$ = 128.4N = $\dfrac{m v^{2} } {r}$

c) T = $\sqrt{ T_{vert} ^{2} + T_{hor} ^{2} }$ = 375.4N

d) The net force on her is simply the tension

e) r, now this is a tricky part

r = 0.28v²

I can't exactly think of a value of 'r'
Title: Re: stonecold physics thread.....
Post by: lakersfan_10 on January 24, 2010, 01:57:12 pm: these are the same question im doing atm. Are you using Nelson Physics?
Title: Re: stonecold physics thread.....
Post by: QuantumJG on January 24, 2010, 02:16:17 pm: ^ did you solve part e?
Title: Re: stonecold physics thread.....
Post by: lakersfan_10 on January 24, 2010, 02:29:50 pm: no i could only get up to part D too
Title: Re: stonecold physics thread.....
Post by: QuantumJG on January 24, 2010, 03:36:54 pm: Quote from: lakersfan_10 on January 24, 2010, 02:29:50 pm
no i could only get up to part D too

what is the answer to 'e'?
Title: Re: stonecold physics thread.....
Post by: stonecold on January 24, 2010, 04:19:25 pm: thanks guys, but i have given up on physics. i need bio more because it will help me with med in the future hopefully, and there is no way i can manage six subjects.

more than happy to give this thread to you lakersfan. just let me know and i'll change the name. and yeah, i 'was' using nelson physics.
Title: Re: stonecold physics thread.....
Post by: lakersfan_10 on January 24, 2010, 04:52:14 pm: its a sad day when someone gives up on physics.
Good luck with bio
Title: Re: stonecold physics thread.....
Post by: QuantumJG on January 24, 2010, 04:58:35 pm: Quote from: lakersfan_10 on January 24, 2010, 04:52:14 pm
its a sad day when someone gives up on physics.
Good luck with bio
Title: Re: stonecold physics thread.....
Post by: stonecold on January 24, 2010, 05:12:27 pm: thanks guys. i'll make this a random question thread for anyone.
Title: Re: random physics questions.....
Post by: lakersfan_10 on January 25, 2010, 01:46:45 pm: OK for part E i got:
Net Force=mass x (speed)^2/radius
131.03=36 x (2)^2/radius
131.03=144/raduis
radius=1.09m

btw for all my previous calculations i took g=10ms