Max Angle of Rod w/ Clay Ball: Rotational Problem 2

[Anthonyphy2013]

Homework Statement

A 75 kg , 34 cm long rod hangs vertically on a frictionless , horizontal axle passing through its cener. A 15 g ball of clay traelling horizontally at 2.2m/s hits and sticks to the very bottom tip of the rod. To what maximum angle , measured from vertical , does the rod ( with the attached ball of clay ) rotate ?

Homework Equations

KE= 1/2 (m+M)V^2 , U= Mg(L-Lcosθ) , mvi=(m+M)(L/2-L/2cos)

The Attempt at a Solution

Energy is conserved , Ei=Ef. Ei= 1/2(m+M)v^2 , Ef = Mg(L-Lcosθ) +(m+M)(L/2-L/2cos) or should i consider K rot as well ?

 

[Simon Bridge]

Energy is conserved , Ei=Ef. Ei= 1/2(m+M)v^2

Start from before the collision - is the collision elastic?

Ef = Mg(L-Lcosθ) +(m+M)(L/2-L/2cos) or should i consider K rot as well ?

The kinetic energy left after the collision turns into gravitational potential energy right - so what would be the rotational KE when the rod reaches it's maximum angle?

note: the rod is pivoted through it's center - how does it's grav PE change with angle?

 

[Anthonyphy2013]

inelastic collision . Thats I am wondering the kinetic rotational energy is concluded and I am wondering the kinetic rotational energy is 1/2IW^2 and I = 1/2MR^2 + (M+m)R^2 ? how can I find angular velocity in this case ?

 

[Simon Bridge]

You need to do conservation of angular momentum for the initial collision, use that $\omega=v/r$
So at the instant just before the collision, the mass m (given) is moving at speed u (given) a distance r (given) from the pivot (the rod has length 2r) then it's moment of inertia is $I_m=mr^2$ and it's momentum is $L=(mr^2)(u/r)$ ... afterwards: the instant after the collision, the mass m is moving with a (different) tangential velocity v, which gives angular velocity for both the mass and the rod (since they are turning at the same rate). Note: moment of inertial of the rod is not Mr^2. Look it up.

The kinetic energy will be $\frac{1}{2}I\omega^2$