- #1
Nicci
- 23
- 0
- Homework Statement
- A particle P of mass m slides on a smooth horizontal table. P is connected to a second particle Q of mass M by a light inextensible string which passes through a small hole O in the table, so that Q hangs below the table while P moves on top. Investigate motions of this system in which Q remains at rest vertically below, while P describes a circle with center O and radius b. Show that P moves with constant speed u, $$u^2 = (Mgb)/m$$
- Relevant Equations
- For particle Q: T - Mg = 0
Thus, T = Mg
For particle Q:
The resultant force on particle Q would be zero since it is at rest. Thus$$ T - Mg = 0$$ which gives $$T = Mg$$
For particle P:
This is where I am struggling. I can't seem to write out the polar equations of motion. I have to show that $$u^2 = (Mgb)/m$$
I know that $$\vec a = (\ddot r-r(\dotΘ)^2) \hat r + (r\ddotΘ + 2\dot r\dotΘ)\hatΘ$$
There has to be two polar equations of motion for P. Should I assume that $$\ddot r= 0$$ and $$\dot r = 0$$ or am I missing something? Then when I say that r=b I will get $$\vec a = (0-b(\dotΘ)^2) \hat r+0$$
I am not sure if I am even on the right track. Can someone please provide me with a hint or some help?
Thank you very much.
The resultant force on particle Q would be zero since it is at rest. Thus$$ T - Mg = 0$$ which gives $$T = Mg$$
For particle P:
This is where I am struggling. I can't seem to write out the polar equations of motion. I have to show that $$u^2 = (Mgb)/m$$
I know that $$\vec a = (\ddot r-r(\dotΘ)^2) \hat r + (r\ddotΘ + 2\dot r\dotΘ)\hatΘ$$
There has to be two polar equations of motion for P. Should I assume that $$\ddot r= 0$$ and $$\dot r = 0$$ or am I missing something? Then when I say that r=b I will get $$\vec a = (0-b(\dotΘ)^2) \hat r+0$$
I am not sure if I am even on the right track. Can someone please provide me with a hint or some help?
Thank you very much.