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Shortening the radius and its effect on speed 
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#1
Mar1805, 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. 


#2
Mar1805, 09:44 AM

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P: 12,016

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? 


#3
Mar1805, 10:04 AM

P: 4,006

Think of angular momentum L. If L is constant then you know that the speed increases as the radius lowers...
marlon 


#4
Mar1805, 03:01 PM

P: 26

Shortening the radius and its effect on speed
As for using momentum, I have no clue as how that will help me. 


#5
Mar1805, 05:28 PM

Sci Advisor
P: 412

{Initial Angular Momentum Magnitude} = L_{0} = m*v_{0}*R_{0} = = (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 L_{0}} ::: ⇒ v = L_{0}/(m*R) {String Tension} = m*v^{2}/R = m*{L_{0}/(m*R)}^{2}/R = (L_{0})^{2}/{m*R^{3}} Thus, the string tension required to maintain circular motion will increase with decreasing radius. The minimum achievable radius "R_{min}" will be that just requiring (25 N) tension force, after which the string breaks: (25 N) = (L_{0})^{2}/{m*(R_{min})^{3}} (25 N) = (1.0332 J*sec)^{2}/{(0.280 kg)*(R_{min})^{3}} = (3.8125)/(R_{min})^{3} ::: ⇒ (R_{min})^{3} = (3.8125)/(25) = (0.1525) ::: ⇒ R_{min} = (0.5343 m) ~~ 


#6
Mar1805, 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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