A ball on a string

  • Thread starter Mewniew
  • Start date
  • #26
15
1
Show me how you get that 5/7 and 2/7 sharing

My notes: https://dl.dropboxusercontent.com/u/34732003/IMG_20150212_180625.jpg [Broken] (dropbox, because its too large)
Where did i go wrong?
 
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  • #27
311
23
Any solid homogenous sphere that rolls without slipping will have the LInear KE and rotating KE in that ratio regardless of speed.

Take the problem from the start (work backwards)
I get the velocity of ball 2 at 120 ° (slack point) to be 2.236 m/s, giving a KE of 25 Joules
Its PE gained at that point = 10 * 10 * 1.5 = 150 Joules
So its KE at collision point must have been: 25 + 150 = 175 Joules
(velocity = sqrt ( 175 / ( ½ * m ) ) = 5.9161 m/s)
This is also the linear KE transferred from ball 1 at the collision
(equal mass, elastic collision)
This represents 5/7 of the total KE of ball 1
( 2/7 is tied up in rotational KE )
So ball 2 total KE ( linear + rotational ) at impact = 175 * ( 7 / 5 ) = 245 Joules
So if m*g*h = 245, then h = 2.45 metres
 
  • #28
311
23
PS: Signing out until: 12.00 noon GMT Saturday
 
  • #29
15
1
Right, got it.
https://dl.dropboxusercontent.com/u/34732003/IMG_20150213_220656.jpg [Broken]. Happened to do another exercise today that helped me a bit with this one.

Took a look at my earlier notes, damn i must've been tired. I think i need a break :F
 
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  • #30
311
23
Glad to help, good for me also.
Dean
 

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