Thanks for your response. I can see now that my calculations were very naive! I had based my calculations on the equivalent of a 2kg particle on the end of a string as you thought.
From your help, I have recalculated the moment of inertia, assuming the reel to be a cylinder with a height of...
Hi everybody,
I'm making a machine at home which will cut a predetermined length of ribbon from a reel. I've designed it all but I'm not sure what sort of spec motor to use.
I'll be using a microcontroller to take the users input and a sensor will sense the size of the wheel. The...
I've now found that the heat capacity ratio is a constant and for this question is 1.4. I don't know why this wasn't mentioned anywhere in the question or my textbooks but a lot of googling came up with the answer in the end.
By using gamma=1.4, T2 comes out as 650K. :)
Homework Statement
An ideal Brayton cycle has a pressure ratio of 15 and can be analysed using air standard cycle assumptions. The gas temperature is 300K at the compressor inlet and 1500K at the turbine inlet.
The compressor and turbine can be considered to be isentropic.
For an...
I've got it now, I couldn't get my head around the difference between m+m and 2 m on the bottom. With the new velocity I get the kinetic energy after the collision to be:
Ek=\frac{1}{2}.m.\frac{2.g.h}{4}
Therefore, Ek=\frac{1}{4}.m.g.h
So the proportion of Ek lost is \frac{3}{4}
Thanks for...
By squaring both sides it works out as:
V = \sqrt{g.h}
I'm pretty sure the units cancel here.
This means that for b) :
The new kinetic energy is \frac{1}{2}.m.g.h
As the kinetic energy changed from m.g.h to \frac{1}{2}.m.g.h, the proportion of energy lost during impact was half.
OK, so if v = \sqrt{2.g.h} at collision,
Ball A's momentum =m.\sqrt{2.g.h}
Ball B has no momentum, so the total system momentum = =m.\sqrt{2.g.h}
Therefore m.\sqrt{2.g.h} /2m = the velocity after the collision
V = \sqrt{2.g.h} /m
For b) Putting the new velocity back into the kinetic...
Homework Statement
Each equal ball shown in figure 1 is suspended on a separate inextensible string of equal length R. The strings are attached to a rigid support and the attachment points are separated by a distance equal to the diameter of the balls. Both balls are initially stationary in...