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M1 pulleys

https://3b0a7b1bc87f5381e60f8f717510b7e3072e9617.googledrive.com/host/0B1ZiqBksUHNYcEVTUFdwYmtsb2c/January%202001%20QP%20-%20M1%20Edexcel.pdf

question 3b, is it because the string is inextensible(doesn't stretch) thus acceleration is the same?
Reply 1
Avatar for B_9710
B_9710
18
They are attached to the same rope So the speed and acceleration of B is the same as the speed and acceleration of A (of course I mean magnitude of acceleration).
Try and visualise it in your head if one of the particles was accelerating more than the other, the rope would either become slack or it would extend (yes you're right the rope is inextensible).
(edited 7 years ago)
Reply 2
Avatar for poundsoffat
poundsoffat
OP
2
Original post by B_9710
They are attached to the same rope So the speed and acceleration of B is the same as the speed and acceleration of A (of course I mean magnitude of acceleration).
Try and visualise it in your head if one of the particles was accelerating more than the other, the rope would either become slack or it would extend (yes you're right the rope is inextensible).


so it's because they're also attached to the same string?
Reply 3
Avatar for KarunN
KarunN
1
http://www.thestudentroom.co.uk/attachment.php?attachmentid=61510&d=1228425984

mark scheme says it's because its inextensible
Reply 4
Avatar for Zacken
Zacken
22
Original post by poundsoffat
so it's because they're also attached to the same string?


It's because the string is inextensible.
Reply 5
Avatar for poundsoffat
poundsoffat
OP
2
Original post by Zacken
It's because the string is inextensible.


oh ok thanks
Original post by poundsoffat
https://3b0a7b1bc87f5381e60f8f717510b7e3072e9617.googledrive.com/host/0B1ZiqBksUHNYcEVTUFdwYmtsb2c/January%202001%20QP%20-%20M1%20Edexcel.pdf

question 3b, is it because the string is inextensible(doesn't stretch) thus acceleration is the same?


Yes, and you can derive the result formally. If the displacement of one mass in a certain time is s1s_1, and the displacement of the other is s2s_2, then, since the length of the the string, ss, is a constant, then we must have:

s1+s2+s=s⇒s1=−s2⇒s1˙=−s2˙⇒s1¨=−s2¨s_1 + s_2 + s = s \Rightarrow s_1 = -s_2 \Rightarrow \dot{s_1} = -\dot{s_2} \Rightarrow \ddot{s_1} = -\ddot{s_2}

i.e. a1=−a2a_1 = -a_2 - the acceleration vectors are the same size, but point in "opposite" directions (I've put "opposite" in scare quotes, since the same result applies regardless of the angle between the two sides of the string e.g. you may have one mass on a slope of a wedge, and the other hanging down its side)
(edited 7 years ago)

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