To calculate v_y let (x,y) be the coordinates of the right end of the rod and let x' be the x coordinate of the left end of the rod - the y coordinate of the left end is always 0.
Now write the function giving y as a function of x'.
I'd have to sit down and work out what that function is, but I doubt it's very hard to derive.
The pressure in a liquid is related to the work done to move a unit mass of the liquid. For example we know the pressure at a depth h is ρgh, and gh is the work done to move a unit mass a distance h.
The work done is ∫F dx and when moving vertically F = mg which is just g for a unit mass. So when moving vertically the integral gives us W = gh.
If we start at the centre and move along the surface the pressure doesn't change. It's just 1 atm everywhere on the surface.
So that means if we move horizontally outwards then vertically upwards, the pressure increase as we move outwards must be the same as the pressure decrease when we move upwards.
Remember we are only calculating the work because the pressure change is proportional to the work done.
So what we are really saying is the horizontal and vertical pressure changes must be the same.
@sanya I've done the calculation this way because you will see this concept used in other similar questions. e.g. suppose we have a tube of length 𝓁 filled with liquid of density ρ and we rotate the tube horizontally with angular velocity ω then what is the pressure at the end of the tube?
At time zero the spring is at its shortest length i.e. extension = 0. Then as we pull the spring stretches to a maximum extension x = 𝓁, then contracts again to its original length x = 0, and this cycle repeats.
This is just like if we have a mass on a vertical spring that is unstretched and we let go. The mass falls down then bounces back up and oscillates vertically in the same way our system above oscillated vertically.
And the key fact here is the maximum extension is half the equilibrium extension.
That is, if we let the mass down slowly so it came to rest with the spring stretched by some distance 𝓁 then that distance would be half the maximum extension.
Now suppose in the system above we ramped up the force slowly so the masses didn't oscillate but just accelerated together. The acceleration would be a = F/m = 90/(3 + 15) = 5 m/s². Yes?
In this question a 90N force is acting on the 15kg mass.
Suppose you were holding the 15kg block stationary on a planet where g = 6 m/s² then the force you would need to hold the block would be the same 90N. Yes?
So physically the motion is the same as if you were on a planet with g = 6 m/s² and you had the 3kg mass on a spring below the 15kg mass you are holding, and then you let the 3kg mass go so it starts bouncing.
The extension of the spring with time is exactly the same as when pulling the system horizontally. It's like you have a horizontal gravity of 0.6g acting.
So in this case the "gravity" force on the 3kg mass is F = m × "g" = 3 × 6 = 18N
Problem Solving Strategies
Problem Solving Strategies