everybody just know off psychology physics here we've got ourselves a national problem involving some tangential quantities some triple quantities look at some of the stuff so here's the system we're gonna take ourselves a little point Neffs attach it to some string some light string we go kind of consider its mass a distance of 0.75 meters away from that to the rotation and we're gonna begin to rotate it with a constant and your acceleration of a 1.5 or radians per second squared we're gonna have some stuff about it so here's a problem you all right so here it is we've got a mass 0.35 zero kilograms the radius at which it is rotating about the axis of rotation the and your acceleration which is constant on 2500 readings per second squared it's initially at rest and we are asked how long does it take this object this mess to have a speed of 12 meters per second now that is a tangential speed how fast it's moving through space so Part A we just go ahead and get to it right we want to know when is B equal to 12 meters per second that's the tangential speed there so when does that occur well again we've got equation of motion for cuts and anger acceleration and we can either deal with tangential quantities or angular quantities in this particular case were already given that your acceleration let's do some angular quantities just because it's there we've got ourselves that Omega final is equal to alpha t plus Omega initial in general and then we've also got ourselves that V sub T is equal to R Omega so we can go ahead and get what Omega final is in terms of what our final tangential speed is so we've got ourselves that Omega final is equal to V divided by R which is our 12 meters per second divided by 0.75 meters [Music] and yes of course that is a 1616 point zero meters caches that we left of per second radians pops up because this is an angular quantity and we've got radians per second so there's the equivalent angular speed with respect to the tangential speed being twelve meters per second so now it's just a matter of when does that occur in time what point of time we're gonna go ahead and take this solve this up for time and get at the time this equal to Omega final minus Omega initial divided by Alpha Omega initial is equal to a zero so we are just left with sixteen point zero radians per second divided by one point five radians per second squared which gives us a grand total cut ten point six six seven seconds ten point six six seven seconds this is the time at which this little mass here will be moving meters per second all right now it's fancy let's continue onward and see what helps we can figure out where Yardena system so got a Part B here we want to know where is the centripetal acceleration after it is rotated around ten revolutions so Part B is the magnitude of the centripetal acceleration act Delta Theta is equal to ten revolutions well we know that this interpretative acceleration is just written a couple different ways just write it like this is our Omega squared or V squared over our we use our Omega squared so there is the magnitude centripetal acceleration it depends on radius and angular speed of this object so the real question then becomes what is in your speed after 10 revolutions so we note that 10 revolutions is equal to 20 pi radians right 2 pi radians per revolution so we've got this and we've also got that Omega final squared is equal to Omega initial squared plus 2 alpha theta the equations of motion for constant angular acceleration which we can then go ahead and utilize to figure out there's something squared right there you know make an initial squared is that's 0 so really Omega final squared is is 2 Alpha Delta Theta which is this right here so in terms of revolutions we could write then that the magnitude of pace of C is equal to R times Omega squared which is 2 alpha don't they we're just pretty listen there with Omega initial is equal to zero we can write out the centripetal acceleration as as such this point of you is that so we want to find this out right we've got ourselves with the magnitude is that equal to 0.75 meters times 2 times 125 bra radians per second squared multiplied by 20 pi radians which leaves us with a grand total for the centripetal acceleration [Music] one forty one point three seven two one forty one point three seven two units of that type of quantity here so that is the inward acceleration that must be correlated with this object's continual change in the direction when it is traveling at the rate that is substantiated by rotating around 20 pi radians under the action of an angular acceleration of 1.5 radians per second squared all right that was a good one looks like there's still a bunch of stuff here that's what's the total acceleration at T is equal to 13 seconds so what do we know about the total acceleration we know that the total acceleration is the vector sum of centripetal and tangential which are perpendicular to one another so really we've got this the magnitude of the total acceleration is equal to the square root of the sum of the squares of a sub T and a sub C so that is the magnitude of the total acceleration vector and we've got the direction of that total acceleration vector arc tangent a sub T over a sub C and that is with respect to the radial inward direction so this is what we're asked for magnitude direction of the total acceleration vector act t is equal to the 30 seconds so what do we need to know a sub T well you know that a sub T is just going to be equal to our times alpha that's all that the tangential acceleration is is R times the angular acceleration so that's constant the time periods is in here because it says R Omega squared and Omega varies as alpha t plus Omega initial so let's we'll write this down a little bit different here and write this out then as this is going to be equal to R times alpha quantity squared that's a sub T and then we've got ourselves R Omega squared R Omega squared quantity squared all right couple little steps here so if we got again that Omega itself is equal to alpha T plus Omega initial this is like an Omega final right well that is Omega as a function of time so this really belongs in there so in this particular case probably easier just to quantify what Omega final it is and stick it in so we can write that Omega final at t is equal to 13 seconds is equal to our 1.5 radians per second squared multiplied by 13 seconds plus 0 and we get that final angular speed of this object after 13 seconds it's nineteen point five radians per second and then we can go ahead and quantify the rest of the stuff that we need so insects all we need to go back to this kit with a sub C AC T is at that point in time so we know if that a sub T is constant which is R times R times out them just 0.75 PI by 1.5 years once per second breaking this up a little bit here meters per second squared and then we've got a sub C is equal to R Omega squared which is equal then to R to 0.75 meters times Omega squared this is going to be Omega at T equals 13 seconds five nineteen point five our readings per second quantity squared squared to e five point eight eight meters per second squared so something just to point out here is of some relevance the centripetal acceleration is greatly greatly greater than the tangential acceleration Y tangential acceleration cousin is always one point one two five meters per second squared if we have a constant alpha in a constant R but some triple acceleration is continuously increasing because of the angular acceleration increase team that may be your velocity as time goes on so then they start up to be around the same value and our small amounts of time but as time continues onward generally speaking the centripetal acceleration x' value overtakes the tangential acceleration value so most of the acceleration is inward as the object speeds up more and more and more at any rate we've got this for the tangential and centripetal acceleration values here so we can go ahead then and get ourselves with that magnitude of the total acceleration then 1.125 m/s squared quantity squared plus two point one two eight two eight five point one eight eight meters per second squared quantity squared yeah that's gonna come out really close to two eighty six I'll just put it in just for argument's sake here quantity squared square root twenty five point one nine what 1/9 of meters per second squared so really if we just branded this off did approximately the same thing it's a little bit more than just in centripetal acceleration there is the magnitude of the total acceleration at T is equal to 13 seconds what direction is this acceleration but ourselves that theta a total arctangent of the tangential divided by the centripetal it looks like it's gonna be really close to zero because it's mostly radially inward but we've still got ourselves our tentative 1.125 and that's about point two to six degrees all right so overall we got the magnitude and direction and this again is with respect to the radially inward direction with respect to Asia HCC's use central focus operation alright that was some fun stuff right there took a little bit of an unraveling but too bad right just keeping track of the tangential versus central quantities oh it will be back to equations of motion for constant angular acceleration that's a great so what is move on then Part D asks what constant velocity which substantiate a centripetal force of 170 Newtons ultimately that's what it asks so we go ahead and proceed with this if magnitude of s of C max is 170 Newton's that is if the maximum central force that this string can withstand or build up to substantiate as 170 Newton's what is the maximum speed tangential speeds at this mask and hat so this is only looking in terms of centripetal little differentiation here momentarily but let's just answer this right so what do we got for this is a triple force and that's what can be written as MV squared over R so you can solve this up for V right it seems pretty straightforward look at ourselves that B then is going to be to the square root of F sub C R over m which is then equal to the square root of 170 Newton's 0.75 meters divided by mass of 0.35 zero kilograms and we get ourselves some V Max speed through space substantiate and there's 70 Newton centripetal force acting on this mess what somebody by urselves 1908 1908 six meters per second so once the mass it's moving that best that is the centripetal force the inward force that must be acting on it to continuously change its direction and ultimately produce this circular path that is maintained so we are asked in Part II what time does that occur at what time does this particular speed a curve which is ultimately asking at what time is the centripetal force 170 so there's a couple different ways that we could go about this again let's use some to some of the tangential quantities right we've got ourselves that V final is equal to a sub T times T plus T initial final tangential velocity is equal to the tangential acceleration times time plus the initial tangential velocity certainly this is just an old equation of motion that we've used before same specifically tangential now we have some stuff here and what do we get out of this what we need to know for one the potential initial is equal to zero objects initially at rest the initial angular velocity is there this initial tangential velocity is zero so we need to know a nice of T is right V sub T is equal to R times alpha which we've already computed just do it again this is going to be equal to 0.75 meters 5 5 is 0.25 meters per second squared so there's the tangential acceleration and well here is our final velocity beat potentials the final is to be 19.0 86 meters per second so we'll just expression right here for tangential velocity as a function of time we sold it out for time we know that this is equal to zero and we've got ourselves the time at which that occurs is then going to be T which is then 18.086 per second divided by at one point one two five meters per second squared sixteen point nine six seconds so we start this object from rest spinning faster faster faster and at some point in time will after sixteen point nine six five seconds some triple force exerted on that mass via the string the attention in the string right is going to be equal to a hundred and seventy news so here's one last question what if talk about the total force acting on this particular mass and say that it's actually the total force that can only build up to be built up to only the under 270 Newtons what does that occur what is the total force on our sub canoes well just like the magnitude of the total acceleration is the vector sum of the tangential instant ripples so is the total force we've got some tangential force which is responsible for the angular acceleration and we've got some centripetal force which is responsible for the centripetal acceleration so we can really resolve this down into looking at those accelerations the total acceleration itself and extracting out at what time that substantiates correlated it's a triple excuse me a correlated total force of this so let's do this so this is just to get the centripetal force should find if we add in the tangential force is responsible for the tangential acceleration the total force being 170 Newtons will occur at a time less than this I was probably not going to be a whole lot less but it's going to be a little bit less let's see how many significant figures we get here so when is the total force equal to this well let us first go ahead and say that the total force the magnitude of the total force is equal to the magnitude the total acceleration were to say M times the magnitude of the total acceleration what are you this well because it's going to be easier just to look at the total acceleration and so with this we've got ourselves 170 Newton's magnitude of the total force divided by zero point three five zero kilograms is equal to the magnitude of the total acceleration that correlates once again with the total force being divided by point 3 5 gives us 45 point 7 1 or 4 85 7 right so there's the total acceleration correlates with this being the total force anyways really close does some other centripetal acceleration that we're looking at all right then so we want to know when is this a correct T equals what regarding that look in the magnitude of the total acceleration to the square root excuse me some of the squares up the tangential it's a triple the tangential is constant the centripetal has time ferry incident what happens if we does go ahead and start making some correlations here I'm gonna make life a little bit easier to not do this completely symbolically is having some numbers here will be just helpful in this particular case so we got forty five point seven one four meters per second squared is going to be equal to the square root a sub T we've already figured out that that is one point one two five meters per second squared that's just our x alpha and then we've got a sub C well a sub C is going to be equal to R times which is pro mega squares right now quantity squared all right so this is what we've got to working with right now let us go ahead and uh we start trying to isolate this because this is what has the time in it right Omega it has the time very isolated Omega and then we can figure out what time I don't make it occurring yet so when it's square both sides so we can write this as the quantity of forty five point seven one four meters per second squared quantity squared is equal to one point one two five meters per second squared quantity squared plus R Omega squared quantity squared right that from both sides and which isn't gonna change this by that much but let's just go ahead and do it right so with this leave this episode we comparison I'm gonna take this value and I'm gonna take this value and let's do this rewrite things a little bit we can write this as R Omega squared this is quantity squared per second squared squared minus five years per second squared squared we could take the square root of both sides and that would be good take the square root of both sides and then we can divide by our good again so we can write this that Omega squared is going to be equal to the square root of this for 85714 dropping the units right now they're completely 0.125 that whole quantity divided by R is equal to Omega squared so let us I'll just quantify that I think that's going to be the easiest thing to do at this point having too much craziness to put in the calculator I'm gonna get a value for that we've got the square root - a bunch of five squared / this comes out to be 647 6 1 76 77 3 all right and actually Radian squared per second squared yes squared per second squared meters cancels out ratings ups in and then we can take the square root of that giving this then Omega is then equal to twenty five point four four feet free I'm handing out some more significant figures we want to see how close we can really get here radians per second and how do we get T this we know that Omega final is equal to alpha T right plus Omega nishal Omega initial was equal to zero so to find this by alpha giving us the T is then equal to Omega divided by alpha which is the twenty five point four four eight three three divided by 1.5 per second squared or radians per second squeezing it in [Music] sixteen point nine six five comes if we're looking at problems where a fatigue is not much much to us than T sub C you get that there is some very big difference between the time at which the total force and the centripetal force are the same values so I'm gonna be careful with that this particular case enough times gotten by does a good problem right I agree alright have a good one so next