Shortening the radius and its effect on speed


by jaymode
Tags: effect, radius, shortening, speed
jaymode
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#1
Mar18-05, 09:41 AM
P: 26
The problem is:
A small block with mass 0.280kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.820m about the hole with a tangential speed of 4.50m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 25.0N.

Well I am really unsure how to proceed with this question. The way I attempted it was trying to find the tension in the string. For that I used the normal acceleration (v^2/r) times the mass. From there I found at the breaking point the acceleration would have to be like 89.29 m/s^2 for the tension to equal 25.0N. Assuming this was all correct I still don't know how to find how the speed changes with the shortening of the radius.
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arildno
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#2
Mar18-05, 09:44 AM
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You don't need to find the tension, really.
Instead, calculate the torques of external forces acting upon the mass with respect to the hole (conceived as a point).
What does that tell you?
marlon
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#3
Mar18-05, 10:04 AM
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Think of angular momentum L. If L is constant then you know that the speed increases as the radius lowers...

marlon

jaymode
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#4
Mar18-05, 03:01 PM
P: 26

Shortening the radius and its effect on speed


Quote Quote by arildno
You don't need to find the tension, really.
Instead, calculate the torques of external forces acting upon the mass with respect to the hole (conceived as a point).
What does that tell you?
Ok. I still dont think i got it right. I calculated the initial torque by multiplying the initial radius by the force. I found that force by finding the normal acelleration(v^2/r). I then set that torque equal to the the new torque which would be 25r. For my answer I got 0.227m for r, which is what i had before and it was not the right answer.

As for using momentum, I have no clue as how that will help me.
xanthym
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#5
Mar18-05, 05:28 PM
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Quote Quote by jaymode
The problem is:
A small block with mass 0.280kg is attached to a string passing through a hole in a frictionless, horizontal surface. The block is originally revolving in a circle with a radius of 0.820m about the hole with a tangential speed of 4.50m/s. The string is then pulled slowly from below, shortening the radius of the circle in which the block revolves. The breaking strength of the string is 25.0N.

Well I am really unsure how to proceed with this question. The way I attempted it was trying to find the tension in the string. For that I used the normal acceleration (v^2/r) times the mass. From there I found at the breaking point the acceleration would have to be like 89.29 m/s^2 for the tension to equal 25.0N. Assuming this was all correct I still don't know how to find how the speed changes with the shortening of the radius.
Let us use circular orbit center "O" for the rotational (Angular Momentum & Torque) reference point. Then, because the block's velocity is always normal to its position vector "R" relative to "O":
{Initial Angular Momentum Magnitude} = L0 = m*v0*R0 =
= (0.280 kg)*(4.50 m/s)*(0.820 m) =
= (1.0332 J*sec)

When the string is pulled, the force on the rotating block is increased, thereby decreasing the block's rotational radius "R". However, at all times during this process, the Torque applied to the block by string tension "F", given by (Torque τ)=(RxF), will be ZERO (0) since the tension force "F" is always parallel to the block's position vector "R" relative to point "O". Hence, Angular Momentum is conserved and is constant for all radii "R":
{Angular Momentum} = L = m*v*R = {Constant L0}
::: ⇒ v = L0/(m*R)

{String Tension} = m*v2/R = m*{L0/(m*R)}2/R = (L0)2/{m*R3}

Thus, the string tension required to maintain circular motion will increase with decreasing radius. The minimum achievable radius "Rmin" will be that just requiring (25 N) tension force, after which the string breaks:
(25 N) = (L0)2/{m*(Rmin)3}
(25 N) = (1.0332 J*sec)2/{(0.280 kg)*(Rmin)3} = (3.8125)/(Rmin)3
::: ⇒ (Rmin)3 = (3.8125)/(25) = (0.1525)
::: ⇒ Rmin = (0.5343 m)


~~
arildno
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#6
Mar18-05, 05:39 PM
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There is no net torque, jaymode!
It follows from this that the angular momentum is conserved about the rotation center, as marlon and xanthym already have said.
However, I didn't see that you were supposed to find the tension and compare it with the breaking strength; I only thought you were to find the new velocity.

xanthym has given you a solution to this.


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