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.