infinite atwood machine problem
Page 1 of 1 • Share •
infinite atwood machine problem
A peculiar Atwood’s machine
The Atwood’s machine shown in the figure consists of masses m,m/2,m/4,........,m/(2^(N1)). All the pulleys and springs are mass less, as usual
a) Put a mass m/(2^(N1)) at the free end of the bottom spring. What are the acceleration s of all the masses
Remove the mass m= (m/(2^(N1)) [which was arbitrarily small, for very large
N] that was attached in part (a). What are the accelerations of all the
Masses, now that you’ve removed this infinitesimal piece?
The Atwood’s machine shown in the figure consists of masses m,m/2,m/4,........,m/(2^(N1)). All the pulleys and springs are mass less, as usual
a) Put a mass m/(2^(N1)) at the free end of the bottom spring. What are the acceleration s of all the masses
Remove the mass m= (m/(2^(N1)) [which was arbitrarily small, for very large
N] that was attached in part (a). What are the accelerations of all the
Masses, now that you’ve removed this infinitesimal piece?
Pinnaka Admin
 Posts : 36
Join date : 20080520
Age : 31
Location : India
Re: infinite atwood machine problem
n part (a), there is perfect balance at all points, so each mass will have acceleration = 0.
Proof: At the last pulley, there are equal masses on each side, so it will not turn. Now on the next pulley above it, the mass on the left equals the sum of the masses on the right, so it will not turn. This continues all the way to the first pulley.
In part (b), all the string tensions are zero, so each mass will accelerate downward with acceleration = g.
Proof: There is a mass on one side of the last pulley, but no mass on the other side. Therefore, the string around the last pully has no tension, so the last mass falls at acceleration g. Since there is no tension in the last string, there is no downward force on the last pulley, so there is no force in the string supporting it. Therefore, the mass on the other end of that string falls freely with acceleration g. This continues all the way to the first pulley.
It is surprising that an arbitrarily small mass can make such a large difference, isn't it?
Hope this helps!
Proof: At the last pulley, there are equal masses on each side, so it will not turn. Now on the next pulley above it, the mass on the left equals the sum of the masses on the right, so it will not turn. This continues all the way to the first pulley.
In part (b), all the string tensions are zero, so each mass will accelerate downward with acceleration = g.
Proof: There is a mass on one side of the last pulley, but no mass on the other side. Therefore, the string around the last pully has no tension, so the last mass falls at acceleration g. Since there is no tension in the last string, there is no downward force on the last pulley, so there is no force in the string supporting it. Therefore, the mass on the other end of that string falls freely with acceleration g. This continues all the way to the first pulley.
It is surprising that an arbitrarily small mass can make such a large difference, isn't it?
Hope this helps!
Pinnaka Admin
 Posts : 36
Join date : 20080520
Age : 31
Location : India
Similar topics
» the car chase with the big yellow machine and its lady driver
» ATM Machine Withdrawal
» Sewing Machine  MANIFESTATION POSTED
» Repost: Getting taught how to shoot a machine gun.
» Yonanas machine
» ATM Machine Withdrawal
» Sewing Machine  MANIFESTATION POSTED
» Repost: Getting taught how to shoot a machine gun.
» Yonanas machine
Page 1 of 1
Permissions in this forum:
You cannot reply to topics in this forum

