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Introductory Physics Homework Assist
Help- speed
Thread starter klm
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[SOLVED] help- speed
Mass m_1 on the frictionless table of the figure is connected by a string through a pigsty in the tabular array to a hanging mass m_2.
With what speed must m_1 rotate in a circle of radius r if m_2 is to remain hanging at rest?
i know i am suppose to testify what i accept done, simply i actually have no idea how to kickoff this problem. any help would be great.
i know that weight( Fg) and tension are working on M2 and that Fg, tension, and normal force are working on M1.
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Answers and Replies
Well round motion seems to be involved. What do yous know nigh that?
umm.. F=ma, for circular movement F=yard V^ii/r and acc points toward the center
So yous know the force acting on 10001 is the weight of m2. How fast should information technology rotate so the centripetal force balances the weight?
i dont know? i dont understand
i know that speed=distance/time ...only i dont know how to figure this out
good.. now that u know this much.. think ahead. Hint: this centripetal force, that u have written, will deed on which body.. and would exist provided by which forcefulness? You are almost there, believe me, merely get a pen and notebook.. and work it out! :)
speed = dist/fourth dimension .. will incorporate more unknowns (irrelevant for the nowadays trouble)! Well, u have already got a relation involving speed... u urself posted it above.. retrieve on that equation.
well i think the centripetal force will act on M1 and the forcefulness is provided by M2, since it is kind of pulling it down. but i dont understand what i am suppose to do next. i dont know what formula to use..
F=m V^2/r ? so i know the forcefulness interim on it is M1..? so square root (Fx r) = 5? i dont know?
I practically told you lot what you have to practice. I'll make information technology clearer. Equate the weight of m2 with the centripetal force.
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i appreciate y'all trying to help me, but it is pretty obvious i am confused and dont understand what to do
hmm.. here is a good news: all formula reqd. for this problem take already been quoted in the threads higher up, u dont need anymore! :) and ya, 1 more thing.. u r right that, information technology is because of m_2, that m_1 has a tendency to be dragged towards the pigsty.. but it can help (and this approach generally helps while solving dynamics issues).. if u think in this line: m_2 is beingness pulled upward by the tension in the string (and not exactly m_1) and m_1 is being dragged towards the hole by tension in the cord (and non exactly m_2)!
You have m1 which is attached to m2 by some string. The weight of thou2 (i.eastward. [itex] Due west=m_2g[/itex]) is acting on m1 via the string. What you're asked to do is find out how fast you need to rotate m1 so that its centripetal force cancels the weight of the starting time mass thoutwo.
Is that any clearer?
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basically, i m trying to suspension the whole system in two dissimilar (although, related through tension) sub-systems.
okay i think i am understanding. this may exist a dumb question merely i have to make certain, the centripetal force is interim on m1 correct?
saket- i empathize the manner you explained information technology most m2 being pulled up and m1 beingness pulled down. and then does that hateful the tension in both are the same then?
so for m2 , would this be correct or necessary : Fnety= T - W = m V^2/r , T-mg=mV^ii/r
centripetal force, by definition, is cocerned with circular motion, and not linear move. [For linear move, g.(5^2)/r, every bit r tends to infinity.. centripetal force tends to zero!]
Yes, tension in both should be aforementioned -- that comes basically from frictionless table.
i dont mean to interrupt but i need help two plz.
For m_2, its radius of curvature tends to infinity (it can just move up-down => linear move), therefore every bit explained above, centripetal force for m_2 is goose egg! Here it also truthful, because, v = 0 for m_2. (It is at rest, correct?)
i dont mean to interrupt but i need aid 2 plz.
If you lot need help and so start your own thread.
For m_2, its radius of curvature tends to infinity (it tin only motion upward-down => linear motion), therefore every bit explained above, centripetal force for m_2 is zero! Here information technology too true, because, v = 0 for m_2. (It is at rest, correct?)
yes.
ohh haha i should have noticed that! thanks though!
ok so i am all the same confused about where to go from here. i know for m2, Fnety= t - mg = mv^ii/r and then for m1 i know that fnet= m v^ii/r. and that m1 w= m1g, and so fnet= m1g 5^2/r
i dunno -- maybe i g trying to over-help! still, i practice not want u to have any confusion.. when nosotros use the centripetal acceleration formula [viz. (v^two)/r], nosotros are concerned with speed and radius of curvature of the particle in question. Therefore, speed and radius of curvature are different for m_1 and m_2. Do not take 'r' as the radius of curvature of m_2. Information technology is in linear move. 'r' is radius of curvature of m_1 because it is in round motility with radius 'r'.
Well, go through the following carefully, and try to figure out where y'all got confused. (It would be a good thought to mail service whatever confusions u had, later in this thread. Also, try to mail service what u learned. -- These are for ur own benefit, believe information technology.)
m_2 is acted down by strength due to gravity ( = m_2.1000). Merely, it is hanging at remainder => the net force on it must be cipher. Only other force acting could be tension in the string. Therefore, T = m_2.one thousand Consider m_1. In vertical management, weight and normal force abolish out each other. Now, m_1 is in circular motion. Therefore, it must have a centripetal acceleration, given by (v^ii)/r. This acceleration can be provided to it merely through the string. Thus, T = m_1.(v^2)/r. I promise from the above 2 equations u would be able to conclude, the required speed.
no i understand that i remember.. ! i know that m1 and m2 are ii sep. , just they have a relationship with each other b/c m2's tension is pulling on m1. i but dont sympathize how to relate the two to solve the problem.
ok i have some other stupid question- practise weight and normal force always cancel each other out for centripetal force?
saket please don't accept this personally simply I do non think you're helping much. I'll reiterate over again.
The force on m1 is the weight of m2. This has to exist balanced by the centripetal force of m1. Thus the weight of m2, must equal the centripetal force of thousand1.
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