I like this, but a small problem
ehild said:
Yes, The vertial component of the tension balances the weight and the horizontal component provides the centripetal force.
*1 Horizontal component: Tsin{alpha}=mv^2/R
*2 Vertical component: Tcos{alpha}=mg
*3 From the geometry of the set-up, it comes that sin{alpha}=R/L,
therefore
T^2cos^2{alpha} = T^2(1-(R/L)^2)=(mg)^2.
You know the maximum allowed tension, so you can calculate the radius of the circle from this last equation.
Knowing R, you can determine sin(alpha), and with that, the linear speed from equation *1.
ehild
Hi there,
The only thing I think that may be incorrect with your reply is the sine and cosine terms. The tension vector is pointing somewhat upwards and towards the origin in the diagram you provided, and the two component forces are the weight, mg, and the horizontal centripetal force labeled F
CP in your attached diagram. So the vertical forces are just the gravity and sin(alpha) = mg/T, but you have cos(alpha) = mg/T. Cosine is the side adjacent over the hypotenuse, and sine alpha is the side opposite over the hypotenuse and the side opposite to the angle alpha in your diagram is the weight mg, so I think you just accidentally mixed the sine with the cosine. So it should be according to your diagram:
Vertical forces = mg = Tsin(alpha) (which should make sense because if the angle is 0, the forces only act in a horizontal direction)
Horizontal forces = F
CP = Tcos(alpha) (which should also make sense because again if the angle is 0 then the entire tension along the string acts inward only)
and cos(alpha) = R/L where R is the radius and hence is the adjacent side to the angle alpha in your diagram, so you want cos(alpha) = R/L. This too should make sense because if the angle is 0 then the mass is moving in a plane with the origin and the length of the string is the radius of the circle the mass is moving around (cos(0) = 1 = R/L, therefore R = L).
Other than this, your analysis is perfect. Your diagram does show a mass hanging from a fixed point moving in circles with not enough speed to bring the mass up so that alpha approaches 0, but that's what happens when the mass is rotated faster and faster. The mass approaches circular motion in a plane and alpha approaches 0, therefore the centripetal force approaches the maximum tension before the string breaks.
So we have:
F
CP = Tcos(alpha) = TR/L now as the mass is spun faster and faster at some point the mass moves nearly in a plane so alpha is nearly zero leaving almost the entire tension along the string to be caused by the centripetal force only, so Tcos(0) = TR/L or T = TR/L, R = L, so the radius is the string length if the mass is moving in circles in a plane.
Now for "
inutard", first I don't appreciate your condescending reply given that I know what I'm talking about and you are simply guessing, so why not read what I have here and maybe you will learn something. Tangential and angular velocity functions were derived by Newton and Huygens in the latter part of the 17th century. It has been known for all of this time that F = mV
2/r = mw
2r is the centripetal force and is a inward acting force only. We are given that when the Tension reaches some value the string breaks and furthermore that when this happens the mass is moving in circles in a near horizontal plane (this isn't explicitly stated but the weight of the mass is 2.94 N and the tension the string breaks at is 25 N, so there's more than enough centripetal force to spin the mass in a near plane), so the tension that breaks the string is virtually the centripetal force acting on the mass through the string. If the mass were rotating slowly enough so that it hung kind of low while moving then gravity would apply a force on the mass that would also act along the string to the center of the circle (like in the attached diagram echild has on his reply), but if the mass is moving in circles in a near horizontal plane the 25N force that breaks the string is almost entirely the centripetal force, as the weight acts to pull the mass down but has virtually no effect on how the mass is pulling on the origin with the centripetal force it has; because the weight of the mass is virtually canceled from the lift it gets from the vertical component of the tension acting along the string. That is why gravity really is not an issue in finding the tangential or angular velocity.
The mass itself only weighs 2.94 N, so if the tension in the string reaches 25 Newtons, there is no doubt that is enough force to keep a 2.94 N mass moving in a near circular plane while rotating. The centripetal force need only overcome the weight of the mass to lift it nearly into a circular plane therefore if the tension reached 25 Newtons the origin feels that force just before the string breaks. The force upon the origin is due almost entirely to the centripetal force so 25 N = mV
2/r = mw
2r. Gravity will always pull down on the mass so the inward force should always be a function of the angle between the mass and the plane the origin sits in. As the speed of the mass rotating is increased it acts to "lift" the mass upwards. When it is first starting to rotate the mass hangs down quite a bit and therefore the tension applied upon the origin from the mass has much to do with the weight of the mass, but as the mass is spun faster the centripetal force acts to pull the mass inward, but since its hanging a little low that inward force can be separated into a horizontal and vertical component, the vertical component acts to lift the mass so that it spins more or less around the origin in the same plane the origin is in.
It cannot lift it to a perfect none zero angle alpha because the closer alpha gets to 0 the greater the vertical component of the string tension needs to be in magnitude to keep the mass lifted and spinning within a plane the origin lies within, but again it cannot lift it entirely into that plane as the angle alpha then is zero and that means there is no vertical lift on the rotating mass, which would cause it to drop a little where there is a vertical component to the tension along the string that acts to keep it nearly in the same plane as the origin; it comes very close to spinning in the same plane as the origin as the centripetal force is bout 10 times the weight of the mass.
So is the angular velocity and tangential velocity independent of gravity? In the strictest sense, no, but for all practical purposes gravity can be ignored and the results are almost identical.
Most periodic circular motion problems do not involve gravity, typical examples may be the moon orbiting the Earth. Of course this is a perfect example of the gravitational force between the two bodies of matter being equal to the centripetal force between them, but the point is there are no other differently directed forces acting on the moon, so the analysis is simply the force between the two bodies is the centripetal force and from that and knowing the distance between the Earth and the moon one can find how fast the moon is spinning around the Earth and the tangential speed of the moon at some location around its orbit at some point in time.
So in the strictest sense, gravity does have an effect on the values desired, but the effect is negligible, particularly at high rotational speeds. Even for this problem if one accounts for gravity in finding the centripetal force acting on the origin from the spinning mass, and someone else disregards gravity and assumes the entire 25 N is the centripetal force, the difference between the answers is one is 0.993 times the other, or there a 0.7% difference between not using gravity and using gravity to find the centripetal force that yields the values desired. A 0.7% difference is well within any margin of error for actual measured data, and the mathematics we do in physics is for predicting the outcome of an event given a set of initial conditions, so what's important is the ability to use the math to predict the physical outcome of some event. If one answer is 0.7% different from another, but is a far easier method for finding that answer, it more than justifies that tiny difference in the answers.
I had said that gravity was not needed for finding the values desired, and for the most part that is true, but in the strictest sense there is a factor of 0.007 difference between using gravity and not using gravity in finding the values desired. That is so small of a difference that I thought it would be better to tell the original poster of the question that he/she does not need gravity to solve the problem. Why make it more complicated than it has to be when the two answers are virtually identical. If you go through and do the math you'll find there is less than a 7 degree difference between the mass and the plane the origin of the circle lies within (alpha = arcsine(2.94N/25N) = 6.75 degrees) so the centripetal force is:
Horizontal force = centripetal force = F
CP = Tcos(alpha) = T*0.993, so including gravitational effects reduces the centripetal force by a factor of 0.007 times the full 25 N tension the string can handle. Those two answers are so close, it is more than safe to assume gravity has no play in finding the tangential or angular velocity, I mean:
Centripetal force including gravity = 24.83 N
Centripetal force not including gravity = 25 N
Horizontal Radius including gravity = R = cos(alpha)L = 0.993*0.5m = 0.497
Horizontal Radius not including gravity = L = 0.5 m
Centripetal force including gravity = 24.83 = 0.3kg*V
2/0.497
re-arranging and solving for V yields:
V = [(24.83*0.497)/0.3kg]^0.5 = 6.41 m/s
Centripetal force not including gravity = 25 N = 0.3kg*V
2/0.5
re-arranging and solving for V yields:
V = [(25*0.5)/0.3kg]^0.5 = 6.45 m/s
Now does a 4 cm/sec difference in velocity justify going through and having to find and properly use the effects from gravity upon the mass? I guess that depends on who is solving the problem and what accuracy is desired, but personally a 4 cm/sec difference in velocity is so small that it is well within any measured variance one would find if they decided to solve this through experimental means, so I think not.
That's my "4 advanced degrees including my Doctorate" speaking to your condescending mind (I'm reffering to inutard not you echild, in fact I like your reply, just the little boo-boo with the sine and cosine operators was the only thing that stuck out).
Craig
