- #1
stunner5000pt
- 1,461
- 2
have a look at the diagram. Mass on top is called Mass 1 and the mass below is caled Mass 2. The spring has spring constant k. At time t=0 the string is cut and the system falls freely. Neglect air resistance
1) Determine the position of the centre of mass at time t>0. The position should be given as distance Ycm froim the CM position at t=0
Nothing spciel about this. Imagin the whole thing as a big body and the mass falls.
[tex] d_{CM} = v_{1} t + \frac{1}{2} g t^2 [/tex]
and since v1 =0
[tex] d_{CM} = \frac{1}{2} gt^2 [/tex]
2) Determine the acceleration on mass 1 and mass 2 immediately after the string has just been cut
to start with for mass 2
kx = mg (1)
for mass 1 T = kx + mg
but once the string is cut T = 0
kx + mg = 0
mg + mg = 0 from 1
2mg = 0
this is the net force 2mg = ma, thus a = 2g
(is there a flaw in this logic?)
for the mass 2
kx = mg but once string is cut kx = 0 since nothing pulls up
thus mg = ma = 0 a = 0??
once again what's the flaw with this logic??
We now refer to the motion for t>0 to a reference frame that has origin in the CM. Denote this frame the position of the lower mass 2 by z.
[B} determine z as a function 0of t. Consider for values of t such that 2 has not hit the floor. [/B]
Now for the this part I am a bit confused.
Certainly z = L/2 + something
what is this something
the text gives the something to be mg/2k cos(root 2k/m) t
not quite sure how they got that part
Thank you in advance for ANY help!
1) Determine the position of the centre of mass at time t>0. The position should be given as distance Ycm froim the CM position at t=0
Nothing spciel about this. Imagin the whole thing as a big body and the mass falls.
[tex] d_{CM} = v_{1} t + \frac{1}{2} g t^2 [/tex]
and since v1 =0
[tex] d_{CM} = \frac{1}{2} gt^2 [/tex]
2) Determine the acceleration on mass 1 and mass 2 immediately after the string has just been cut
to start with for mass 2
kx = mg (1)
for mass 1 T = kx + mg
but once the string is cut T = 0
kx + mg = 0
mg + mg = 0 from 1
2mg = 0
this is the net force 2mg = ma, thus a = 2g
(is there a flaw in this logic?)
for the mass 2
kx = mg but once string is cut kx = 0 since nothing pulls up
thus mg = ma = 0 a = 0??
once again what's the flaw with this logic??
We now refer to the motion for t>0 to a reference frame that has origin in the CM. Denote this frame the position of the lower mass 2 by z.
[B} determine z as a function 0of t. Consider for values of t such that 2 has not hit the floor. [/B]
Now for the this part I am a bit confused.
Certainly z = L/2 + something
what is this something
the text gives the something to be mg/2k cos(root 2k/m) t
not quite sure how they got that part
Thank you in advance for ANY help!
Attachments
Last edited: