A rod of length L, whose lower end is resting along the horizontal plane, starts to topple from the vertical position.what will be the velocity of upper end when it hits the ground?

A projectile's launch speed is5X speed at maximum height.Find launch angle 0.

How is projectile related to this problem?

It seems 7318252 has tried to hijack your thread.

I suggest addressing the falling rod problem using energy principles. The kinetic energy of the rod whan it hits the ground is:

\Delta KE= \frac 1 2 I \omega^2

'I' is the moment of inertia, which is dictated by the rod's mass and length. This change in KE equals the loss in potential energy as the center of gravity falls a height of L/2, or

\Delta PE = - mg(\frac L 2 )

Given \Delta KE + \Delta PE = 0 , you can solve for \omega , and hence the velocity of the tip of the rod upon impact with the ground.

A projectile's launch speed is5X speed at maximum height.Find launch angle 0.

Please don't tag an unrelated question onto an existing thread. I see that you also asked this same question here: https://www.askmehelpdesk.com/physics/projectile-motion-753656.html where it has already been addressed.

at ebaines, one question why did you use centre of gravity in this problem and can you tell value of I(moment of inertia) because I am a bit confused with it's value with some explanaton?

1. I used the center of gravity because change in PE is a linear function of the height of the fall, so you can use the average height - assuming that the rod is uniform in mass distribution. For a more formal explanation consider this: for each section of the rod the change in PE is \d(PE) = -\rho g y dy , where \rho = density per unit length and y = original height of each section dL. From this:


\int d(PE) = - \displaystyle \int_0 ^L \rho g y dy \\

\Delta PE = - \frac g 2 \rho L^2


The mass of the rod is m = \rho L, so this becomes


\Delta PE = - \frac 1 2 mgL


As for the value of I, the momemt of inertia - your text should provide some examples, but for a rod that pivots about its end it's I = mL^2/3. For future reference here is a pretty good list :
List of moments of inertia - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/List_of_moments_of_inertia)

You can always calculate I from basic principles, and indeed you should learn this technique:


I = \int \rho r^2 dr


For a rod pivoting at one end this becomes:


I = \ \displaystyle \int _0 ^L \rho r^2 dr = \rho \frac {L^3}3 = m \frac {L^2} 2

Hope this helps.

Got it... thanks for helping...

You have to consider two cases: (a) there is no friction between the rod and the surface -- the lower end only rests initially and then moves; (b) the lower end is always at rest because of the friction. The answers are different.