• Support PF! Buy your school textbooks, materials and every day products Here!

Strangely Confusing mechanics problem

  • Thread starter SelfStudy
  • Start date
  • #1
7
0
Strangely Confusing mechanics problem!!!

Homework Statement



2 masses connected by a massless string which passes through a hole in a horizontal table. One mass, [tex]m_{1}[/tex], is sitting on top of the horizontal table, the other, [tex]m_{2}[/tex], hanging below the table. There IS friction between mass [tex]m_{1}[/tex] and the tabletop, coefficient of friction=[tex]\mu[/tex]. Mass [tex]m_{1}[/tex] has initial velocity [tex]v_{0}[/tex] directed perpendicular to the string.

Find the initial radius of the circular motion, the initial rate of energy dissipation, the speed after time T, and the radius after time T

Homework Equations



Newton's Laws, centripetal acceleration, kinetic energy, etc

The Attempt at a Solution



So, as the top mass moves it loses energy to friction and so it's velocity and radius should decrease, the tension in the rope declines as well so the lower mass descends.

The initial radius is found from a free body diagram and Newton's 2nd law. At first (at t=0) the tension balances the weight of M2 and since the tension is what provides the centripetal force I have:

[tex]T=\frac{m_{1}v_{0}^{2}}{r_{0}}=m_{2}g \Rightarrow r_{0}=\frac{m_{1}v_{0}^{2}}{m_{2}g}[/tex]

For the initial energy dissipation,

[tex]P=f\bullet v=\mu m_{1}g v_{0}[/tex]

However, for the speed and radius t seconds later I can't just multiply the power found above by the time, t, because the velocity (and hence the dissipation rate) has changed, right?. I don't know how far the mass has moved so I can't calculate the work done against friction either.

Can anyone suggest a tip?

Thanks.
 

Answers and Replies

  • #2
phyzguy
Science Advisor
4,483
1,436


Your relationships for T and P continue to hold as r decreases. Since you know v(r), you can calculate the total energy E(r) as a function of r. Differentiating it gives dE/dr. Since you know P=dE/dt, the ratio of these two gives dr/dt, which you can integrate to give r(t). You can then plug that back in to get v(t).
 
  • #3
7
0


Thank you for the response.

I don't quite follow your suggestion though. The relationship
[tex]T=\frac{m_{1}v_{0}^{2}}{r_{0}}[/tex]
holds, but is not equal to [tex]m_{2} g[/tex] because the tension no longer equals the force of gravity on m2. So I don't think I know v(r), and so can't calculate E(r).
 
  • #4
7
0


I'd really appreciate any help on this problem. It's not for homework or anything, I just can't understand where I'm going wrong.

Thanks.
 
  • #5
rl.bhat
Homework Helper
4,433
7


If m1 is not moving in the circular orbit, what will be the acceleration of m1 ( which is equal to m2) taking into account the frictional force .
By knowing the acceleration of m2, you can find the distance traveled by m2 in time T.
There is no tangential force acting on m1. So its angular momentum is conserved.
Hence I1*ω1 = I1*ω2.
With these hints can you proceed with the problem?
 
  • #6
7
0


Thanks rl.bhat

I will try and work it out with your hints, thanks.

However, isn't there a component of friction which acts tangent to the motion of m1?
 
  • #7
7
0


I guess what is confusing me is the fact that m1 moves in a spiral, getting closer and closer to the hole in the table. In this case we can't assume that
[tex]T=\frac{m_{1}v^{2}}{r}[/tex]
Is that correct?

Also, because friction acts in the direction that opposes the velocity, there is always a component of friction that is tangent to the motion of m1 so I can't assume that angular momentum is conserved.

The acceleration of m2 is found from:
[tex]T-m_{2}g=m_{2}a \Rightarrow a=\frac{T}{m_{2}}-g[/tex]

But now I don't know T because I can't just use the form of the centripetal force above (m1 isn't moving in a circle and there is a component of friction that acts in the opposite direction of T).

I'm pretty stuck at this point...
 
  • #8
7
0


Can't figure this one out! Please help!!
 
  • #9
rl.bhat
Homework Helper
4,433
7


You are right. There is an acceleration in the tangential direction. Using this you can find the angular velocity after time T.
Your method of calculation of acceleration of m2 is not correct. It should be
m2g - T = m2*a......(1)
T - Fr = m1*a......(2) Here Fr is the frictional force.
Now solve the equations to find a.
To find the initial radius, the tangential acceleration need not be considered. Only initial velocity and acceleration of m2 is sufficient.
 
  • #10
7
0


Hi,

Yes, my equation for the acceleration of m2 was off by a sign: a should have the same sign as g.

But for Eq (2) i'm a little confused. Newton's 2nd law is a vector equation, so for which component have you written the equation? The force of friction acts at some angle to the tension, correct? So you can write it in terms of a component perpendicular to T and one that points anti-parallel to T. Which means that there should be some angle dependent term attached to the friction force, no?
 
  • #11
rl.bhat
Homework Helper
4,433
7


m1 is moving tangentially and radially.
Tangential frictional force acts in the opposite direction to the velocity. It does not contribute to the tension. Radial frictional force apposes centripetal force which is the tension in the string. Combination of these two causes the m1 to move in a spiral of reducing radius.
 

Related Threads on Strangely Confusing mechanics problem

  • Last Post
2
Replies
29
Views
2K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
5
Views
1K
  • Last Post
Replies
8
Views
1K
  • Last Post
Replies
4
Views
1K
  • Last Post
Replies
3
Views
3K
Replies
3
Views
2K
  • Last Post
Replies
13
Views
4K
Top