Solve Centripetal Force for m_1 in Frictionless Table

[klm]

[SOLVED] help- speed

Mass m_1 on the frictionless table of the figure is connected by a string through a hole in the table 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 show what i have done, but i really have no idea how to start 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.

 

[Kurdt]

Well circular motion seems to be involved. What do you know about that?

 

[klm]

umm.. F=ma, for circular motion F=m V^2/r
and acc points toward the center

 

[Kurdt]

So you know the force acting on m₁ is the weight of m₂. How fast should it rotate so the centripetal force balances the weight?

 

[klm]

i don't know? i don't understand

 

[klm]

i know that speed=distance/time ...but i don't know how to figure this out

 

[saket]

good.. now that u know this much.. think ahead. Hint: this centripetal force, that u have written, will act on which body.. and would be provided by which force? You are almost there, believe me, just get a pen and notebook.. and work it out! :)

 

[saket]

speed = dist/time .. will incorporate more unknowns (irrelevant for the present problem)!
Well, u have already got a relation involving speed... u urself posted it above.. think on that equation.

 

[klm]

well i think the centripetal force will act on M1 and the force is provided by M2, since it is kind of pulling it down. but i don't understand what i am suppose to do next. i don't know what formula to use..

 

[klm]

F=m V^2/r ? so i know the force acting on it is M1..? so square root (Fx r) = v? i don't know?

 

[Kurdt]

I practically told you what you have to do. I'll make it clearer. Equate the weight of m₂ with the centripetal force.

 

[klm]

i appreciate you trying to help me, but it is pretty obvious i am confused and don't understand what to do

 

[saket]

hmm.. here is a good news: all formula reqd. for this problem have already been quoted in the threads above, u don't need anymore! :)
and ya, one more thing.. u r right that, it is because of m_2, that m_1 has a tendency to be dragged towards the hole.. but it can help (and this approach generally helps while solving dynamics problems).. if u think in this line: m_2 is being pulled upwards by the tension in the string (and not exactly m_1) and m_1 is being dragged towards the hole by tension in the string (and not exactly m_2)!

 

[Kurdt]

You have m₁ which is attached to m₂ by some string. The weight of m₂ (i.e. W=m_2g) 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 first mass m₂.

Is that any clearer?

 

[saket]

basically, i m trying to break the whole system in two different (although, related through tension) sub-systems.

 

[klm]

okay i think i am understanding. this may be a dumb question but i have to make sure, the centripetal force is acting on m1 correct?

 

[klm]

saket- i understand the way you explained it about m2 being pulled up and m1 being pulled down. so does that mean the tension in both are the same then?

 

[klm]

so for m2 , would this be correct or necessary : Fnety= T - W = m V^2/r , T-mg=mV^2/r

 

[saket]

centripetal force, by definition, is cocerned with circular motion, and not linear motion. [For linear motion, m.(v^2)/r, as r tends to infinity.. centripetal force tends to zero!]

Yes, tension in both should be same -- that comes basically from frictionless table.

 

[ineedhwhelp]

i don't mean to interrupt but i need help 2 please.

 

[saket]

For m_2, its radius of curvature tends to infinity (it can only move up-down => linear motion), therefore as explained above, centripetal force for m_2 is zero! Here it also true, because, v = 0 for m_2. (It is at rest, right?)

 

[Kurdt]

i don't mean to interrupt but i need help 2 please.

If you need help then start your own thread.

 

[ineedhwhelp]

For m_2, its radius of curvature tends to infinity (it can only move up-down => linear motion), therefore as explained above, centripetal force for m_2 is zero! Here it also true, because, v = 0 for m_2. (It is at rest, right?)

yes.

 

[klm]

ohh haha i should have noticed that! thanks though!

 

[klm]

ok so i am still confused about where to go from here. i know for m2, Fnety= t - mg = mv^2/r
and then for m1 i know that fnet= m v^2/r. and that m1 w= m1g, so fnet= m1g V^2/r

 

[saket]

i don't know -- maybe i m trying to over-help! still, i do not want u to have any confusion.. when we use the centripetal acceleration formula [viz. (v^2)/r], we 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. It is in linear motion. 'r' is radius of curvature of m_1 because it is in circular motion with radius 'r'.

 

[saket]

Well, go through the following carefully, and try to figure out where you got confused. (It would be a good idea to post whatever confusions u had, later in this thread. Also, try to post what u learned. -- These are for ur own benefit, believe it.)

m_2 is acted down by force due to gravity ( = m_2.g). But, it is hanging at rest => the net force on it must be zero. Only other force acting could be tension in the string. Therefore, T = m_2.g
Consider m_1. In vertical direction, weight and normal force cancel out each other. Now, m_1 is in circular motion. Therefore, it must have a centripetal acceleration, given by (v^2)/r. This acceleration can be provided to it only through the string. Thus, T = m_1.(v^2)/r.
I hope from the above two equations u would be able to conclude, the required speed.

 

[klm]

no i understand that i think.. ! i know that m1 and m2 are two sep. , but they have a relationship with each other b/c m2's tension is pulling on m1. i just don't understand how to relate the two to solve the problem.

 

[klm]

ok i have another stupid question- do weight and normal force always cancel each other out for centripetal force?

 

[Kurdt]

saket please don't take this personally but I do not think you're helping much. I'll reiterate again.

The force on m₁ is the weight of m₂. This has to be balanced by the centripetal force of m₁. Thus the weight of m₂, must equal the centripetal force of m₁.

 

[klm]

are you saying that m2= m1 v^2/r

 

[klm]

so m2g= m1 v^2/r

 

[Kurdt]

are you saying that m2= m1 v^2/r

Almost. Remember weight is the mass multiplied by the acceleration due to gravity.

W_2 = m_2g

Can you work out v from there?

 

[klm]

square root (m2xgxr/ m1) = v ?

 

[Kurdt]

square root (m2xgxr/ m1) = v ?

Yes. Sorry I wasn't clearer earlier.

 

[klm]

no i am sorry. i have a hard time understanding physics, and i really appreciate the help you have given me.

 

[Kurdt]

So do you understand how this problem works now more importantly?

 

[klm]

i think so, though i do have one question. how come normal force and weight canceled each other out in m1? does that always happen in centripetal force?

 

[Kurdt]

Well m₁ is resting on a table which means its weight is counter acted by a normal force. That just comes from Newton's 3rd law of equal and opposite reactions. Also the weight and the normal force on m₁ act perpendicular to the centripetal force so they don't really come into the problem.

If I'm interpreting you wrong and you're asking why the centripetal force canceled the weight then simply because that's what the question asked you to do. So you set both equations equal and found out what speed m₁ would have to travel in a circle to balance m₂. Of courseif you were to set this up in a lab and you randomly spun the mass on the table it might be going too fast or too slow to balance the weight of m₂.

If that doesn't answer it then I'm not sure what weight you're referring to and what normal force.

 

[saket]

i think so, though i do have one question. how come normal force and weight canceled each other out in m1? does that always happen in centripetal force?

No, it is NOT a rule of thumb.
I suggest you making a free body diagram (fbd) of m_1. You already have identified the forces acting on it, when seen from ground, as the weight, normal force and the tension. (ur first thread.)
In the fbd of m_1 u will see that, in this problem, in the vertical direction only weight and normal force are acting. But since m_1 is neither lifting up from the table nor breaking into the table, (note that m_1 has to move in the plane of table), acceleration of m_1 in vertical direction is zero. Using Newton's 2nd Law: m_1.g - N = m_1.(0) => N = m_1.g
We left out tension, T, in so far discussion of fbd because it was in a perpendicular to the above two forces and as u very well be knowing that component in a perpendicular direction is FCos90 = 0.
Now in the horizontal direction (ie, along the plane of table), only force is this tension. Therefore, this must provide necessary centripetal force for the m_1 to move in a circular path.

 

[saket]

saket please don't take this personally but I do not think you're helping much.

it's okay sir. i m just a new entry to this forum.. i m still learning how to handle things. go on.. u r recognized homework-helper.

 

[klm]

thank you both very much. i understand now about the forces canceling out.