M1 Pulleys Question

Two particles P and Q of masses 1kg and 2kg respectively are hanging vertically from the ends of a light inextensible string which passes over a smooth fixed pulley. The system is released from rest with both particles a distance 1.5m above a floor. When the masses have been moving for 0.5s the string breaks. Find the further time that elapses before P hits the floor.

Thanx for any help anyone can give

Tut

Reply 1

Swordsman

Use F=ma for both sides of the pulley to get
2g-T=2a and T-g=a
Find a, this is the accelreation that the particle with mass 2g is coming down with. Then use equations of motion to work out the velocity when the string breaks, and the distance at this point. Then work out the distance it still needs to go before hitting the ground and then the time it takes to do this.
There may be a quicker way, but that's how I would do it.

Reply 2

Mohit_C

First of all one moves up and the other moves down. As Q is heavier, Q goes up and P goes down. First let's find the acceleration of the system.
For P:
F=ma
T-g=1*a
T-g=a

For Q:
2g-T=2a

Add both equations:
2g-g=3a
g/3=a

Now let's find where Q and P are when t=0.5
As released from rest, u=0
v=u+at
v=g/3*0.5
v=1.633 m/s.
at this time Q is s=ut+1/2at^2
Q has travelled: 1/2*g/3*0.5^2
=0.40833 metres down, therefore P is 1.5+0.40833=1.9083 metres up (as it goes up when Q goes down).
Now it slows down. So, v=u+at
v=0
u=1.633
0=1.633-9.8t
t=0.1666 seconds to go up to when it stops.
therefore in these 0.166 seconds, it has travelled:s=1.663(0.166)-0.5*9.8*0.166^2
s=0.1361 metres, and so now is 1.9083+0.1361=2.044metres up.

At this instant:
u=0
v=?
a=9.8
so time taken to get down =
s=ut+0.5at^2
2.044=0.5*9.8*t^2
Rearrange and solve to get t=0.646 seconds to get down. But don't forget the time it took to get to 2.044 metres up so 0.646+0.166=
0.813 seconds

I hope i didnt make a mistake.

Reply 3

Codefusion

2g - t = 2a
t -g = a
g = 3a
a = 1/3g
now find the velocity when the string breaks:
u=0
a = 1/3g
t = 0.5
v = 1/3g * 1/2 = 1/6g
s = 1/6g * 1/4 = 1/24g
now find how much time will elapse before it starts falling
a = -9.8
u = 1/6g
v = 0
t = -1/6g / -g = 1/6
s = 1/6g * 1/6 - 1/2g * 1/36 = 2/72g - 1/72g = 1/72g
now the final eqn
a = g
u = 0
s = 1/24g + 1/72g + 1.5 = 92/441g
t = ?
1/2gt2 = 92/441g
1/2t2 = 92/441
t2 = 184/441
t = sqr(184/441)
time = 1/6 + sqr(184/441) = 0.8126s
hope this helps.

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