A 30 g mass is attached to one end of a 10-cm-long spring. The other end of the spring is connected to a frictionless pivot on a frictionless, horizontal surface. Spinning the mass around in a circle at 90 rpm causes the spring to stretch to a length of 12 cm. What is the value of the spring constant?

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Hi everyone. In this particular fact, this problem, we are asked to determine the spring constant of an ideal spring where it will be welded at one end to the so surface of a spin coder. And when it is at rest horizontally on a surface, the spring will have an equilibrium length of 14 cm. And when we're attaching a small object of mass, 55 g to the freon of the spring, the spin coder is rotated at a speed of 1 25 R PM. Once those two are moved, then the spring will move in a circular path at a constant speed and becomes 18 centimeters. We're asked to calculate the spring constant. And first to do this, I am going to list all the given information in our problem statement. So at equilibrium, we have the string uh the length to b 14 centimeters. So there are B length in equilibrium 14 centimeters and the length after it's actually moving Or LF length final is 18 cm. The mass of the object is 55 g Which equals to uh 55 times 10 to the power of -3 kg. And the spin coder is rotated at a speed of 1 R PM, which will be our angular axel uh our angular velocity here W or our omega which is 1 25 R PM. We wanna con uh convert the R PM into rat per seconds. So 1 25 R pm is 1 25 revolution per minute, converting the revolution to ra uh to ra radium. We want to multiply it by two pi Rat over and then we want to multiply it by one minute For 60 seconds. And that will give us an omega of 13. rat per second or the angular philosophy of rat uh 13.1 radiance per seconds. Um I am going to change the or convert the length into meter. So this will be 1.10 point 14 m for the equilibrium length and 0.18 m for the final length. So now we are ready to actually start our calculation. So the small object actually undergoes a uniform circular motion and the spring force actually provides the cent pedal force to maintain the circular path at a constant speed. We want to apply Newton's second law, Newton second law in the radial direction. So that will be Sigma F R equals M multiplied by A in the radial direction or the acceleration in the radial direction, which is A R. So that will be, as uh I've said previously, the spring uh force is going to be what's causing the center pedal force. So the spring force is K multiplied by delta L or the elongation. And uh free deal acceleration can actually be calculated by uh substituting A R with V squared over R or the um radius of the center pedal movement. So V squared over R can then be also calculated or be substituted with R multiplied by W squared. So A R equals V squared over R equals R multiplied by W squared or omega squared. So with this equation right here, we can actually rearrange this so that we can find the spring constant. So the spring constant will equals to M R omega squared divided by delta L. So this is going to be our equation to calculate the spring constant. OK. So um we actually have all this information right here and we can start by actually plugging everything into our equation. So K will equals to our mass which is uh 55 g or 55 times 10 to the power of minus three kg multiplied by our um R or the radius of the centipede movement. The R is going to actually be just the final length of the spring which is going to be 18 centimeters or 0.18 m Multiplied by the omega squared, which is, we have calculated that to be 13.1 rat per second squared. All of that divided by delta L delta L is the elongation of the spring. The elongation is going to be L F minus L E. So delta, L equals L F or final minus L E which is the equilibrium and delta will be 0.18 m minus 0.14 m. And that will give us a spring constant of 42.5 newtons per meter. So a spring constant of 42.5 newtons per meter is going to be the answer to this practice problem and that will correspond to option C. So option C is going to be the answer with a spring constant of 42.5 newtons per meter. And that'll be all for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and they'll be all for this one. Thank you.

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Key Concepts

Hooke's Law

Centripetal Force

Spring Constant