Two-Dimensional Kinematics: Ball on a string

  • #1
Here's the problem.

A physics student whirls a small stone on a string in a horizontal circle. The circle has a radius of 1.4 m and a height of 2.2 m above the ground. Unfortunately, after she has whirled it for 33 seconds, the string breaks and the stone flys off, landing 12 from her (and just barely missing her physics teacher taking a nap near by).

question is:
What was the magnitude of the centripetal acceleration on the stone just before the string broke?

Well, I know that for centripetal acceleration, a=v^2/r
where a=acceleration
r= radius.

The other equation I know is T= (2(pi)r)/v
where T=period

At first I thought this was easy, but after half an hour of frustration, computer won't take any of my answers.

I know that one thing is find what is the velocity for the stone to land 12meters away, but I don't know how to do it. I tried making a triangle where the sides are 1.4m and 2.2m, and solve for the hypotheneuse, which gives me 2.6088 meters, but I don't know what that means really, if anything.

I tried manipulating the two forumulas to no avail, but no luck whatsoever. What am I not getting this time?

Answers and Replies

  • #2
Wow! This is probably one of the best kinematics problems I've ever seen. I haven't solved it yet, but here are my thoughts. To find a, what do you need? You need to find r and v2. You know r. How do you find v2? You need to note that v2=(vsinT)2+(vcosT)2=vx2 + vy2. So it seems no matter what v_x and v_y you have, v will be the same. Now how do you find v using this? Use the information concerning the range.
The rest is up to you.
I don't know for sure about my answer though, because I'm assuming v is constant, and to be honest I don't know that it is.
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  • #3
Um..yea. What do you mean "use the information concerning the range?" Also, I understand v^2=(vsinT)^2+(vcosT)^2=vx^2 + vy^2,but how do I figure out T?
  • #4
Hmm... I made a mistake.
Wait a moment please.

I change my response to not enough information. You need T.
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  • #5
Originally posted by StephenPrivitera
So it seems no matter what v_x and v_y you have, v will be the same.
This is wrong.

Edit: No it isn't. I don't know anymore. I keep going back and forth. I think this is right. If it is, then v does not depend on T.
Use r=(vcosT)tflight
find tflight
and then choose a T that makes the problem easy to solve for v
That's what I did anyway.

Anyone else?
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  • #6
So the important thing to remember here is that this is a ballistic problem and the stone has to travel the horizontal distance (dx) in the same time it travels the vertical distance (dz). So you use the 12m and the 2.2m to find the time to fall and then use that to give you the velocity of the stone. The important thing to remember there is Newton's first law, object in motion will continue in motion unless acted upon by an outside force...the outside force was the string, once it broke the stone had no choice but to fly away in a straight line, and fall to the Earth via gravity. Now, remember that the v in a=(v^2)/r is the tangential velocity, i.e. the velocity of the line perpendicular to the circle of radial motion, and THAT is the velocity that the stone has when the string breaks, see the stone flies off in the straight line.

Hope you're still with me draw a picture with the ground and the stone 2.2m above it and then a "x" on the ground 12m away and connect them with a parabola. Now to the actual solving, I get 916.36m/s^2 by the way. But start with 12m=dx(distance-x)=vx(velocity-x)*t AND 2.2m=dz(z is vertical)=vz*t See about the time being the same?

Now, however you get the equation, you need the velocity-acceleration-distance equation. I may be weird but I start with Newton's second law and integrate. (by the way, Newtons second law actually deals with the momentum differential, F=dp/dt, but that simplifies to F=ma for a constant mass) So F=ma and the force we're interested in solving for is gravity where F=mg so mg=ma cancel the mass, g=a but a=dv/dt derivative of velocity with respect to time or a=(dv/dx)*(dx/dt) replace dx/dt=v gives a=v*dv/dx Substitute that to g=a=v*dv/dx "move" the dx to g*dx=v*dv then think about how we're talking about vz for gravity so g*dx=vz*dv and integrate both sides using appropriate limits. Integrate g*dx from 2.2m initial to 0m final and v*dv from 0m/s initial to vz final and you get the equation v^2=2*a*d evaluated to -2.2*g=(vz^2)/2 which looks wrong until you remember that g=-9.8m/s^2 by convention of pointing down. So vz=sqrt(2*-2.2m*g)=sqrt(2*-2.2m*-9.8m/s^2)=6.57m/s Now that's the vertical velocity for falling from that height.

With this you can get the time, vz=dz/t so t=dz/vz=2.2m/(6.57m/s)=.335s and this is the time for falling down which must equal the time for traveling horizontally, so dx=vx*t gives vx=dx/t=12m/.335s=35.82m/s And this vx is the velocity the stone had when the string broke which was the tangential velocity so it can be used to get the centripetal acceleration a=vt^2/r=(35.82m/2)^2/1.4m=916.36m/s^2

Hope I didn't get too carried away with my explanation!
  • #7
Ok. I tried out your answer, but it didn't work. So I scrutinized your work as best as I could, and you said "And this vx is the velocity the stone had when the string broke which was the tangential velocity so it can be used to get the centripetal acceleration a=vt^2/r=(35.82m/2)^2/1.4m=916.36m/s^2"

a=v^2/r, so of course I thought a=(35.82)^2/1.4 = 916.5 m/s^2
That didn't work.

Then I looked at this line:
"a=vt^2/r" and that didn't make sense.
Anyhow, Zimm put down a=(35.82m/2)^2/1.4m = 229 m/sec^2 and that's the correct answer! Now I'm totally confused! My question is, why is 35.82 divided in half in order to get the correct answer??[?]
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  • #8
Yeah, I had a typo and an error there...the "a=vt^2/r=(35.82m/2)^2/1.4m=916.36m/s^2" had the typo, I meant 35.82m/s

Now for the error, I know it's where I set vz=dz/t You see, the vz I calculated is the speed of the stone when it hit the ground, but that is not the speed throughout the fall, see the average speed is halfway, because the stone starts falling at 0m/s and ends falling at 6.57m/s so the average is 6.57/2 and THAT is what you need to use to get the time, so the two carried the error through the rest of my calculation. So I needed t=dz/vz-average=2.2m/(3.28m/s)=.67s

Now, no such average velocity stuff applies to vx because there is no acceleration in the x-y plane (parallel to the ground, and neglecting air resistance) so dx/t=12m/.67s=17.91m/s=vx then vx=vt and a=vt^2/r=17.91^2/1.4=229.12

I'll see if I can work around my average velocity argument in a little while because the only way that works is in a constant gravitational field where acceleration is constant and there should be a way to account for variable acceleration.
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  • #9
Ok, here I see it, I needed the time, but my main differential F=ma -> m*g=m*a -> g=a -> g=dv/dt gives me time versus velocity, which is why I did the swith to g=(dv/dx)*(dx/dt) -> g=v*dv/dx to give me velocity versus distance, see I knew the distance so I was able to get the velocity. So the next step was to take the velocity and get the time, so you go back to g=dv/dt and get g*dt=dv then integrate both sides, let t go from 0 to t the time to find and v go from 0 to vz that was found in the first part, so the integral turns to g*t=vz -> t=vz/g=(6.57m/s)/(9.8m/s^2)=.67s
  • #10
why average velocity

It sort of all making sense now, but how did you know that average velocity is what I needed to finish the problem? And why is it average velocity?
  • #11
Zimm has his own way of doing things, but I think I may have found an easier way of going about it.

By using this equation... y-y0=voy(t)+(1/2)g(t)^2
where y=0
Solve for t to get 0.6700 sec. Thats the time for the stone to hit the ground in the y direction.

Now you know that Vx=Dx/t => velocity in x direction = 12m/.6700sec=17.91 m/sec...that is the velocity stone had in the x-direction.

So now use a=v^2/r => (17.91)^2/1.4= 229 m/s/s

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