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 $\frac{1}{2}I\omega^2$

 

[Nickie]

I would approach this problem by first identifying the key concepts and equations that are relevant to the situation. In this case, the key concepts include conservation of energy, rotational motion, and the equations for kinetic and potential energy.

To determine the maximum angle of rotation, we can use the principle of conservation of energy. This means that the initial kinetic energy of the clay ball must be equal to the final potential energy of the system (rod with attached clay ball). We can set up an equation using the kinetic energy of the ball and the potential energy of the system, and solve for the maximum angle θ.

We also need to consider the rotational kinetic energy of the system, which can be calculated using the moment of inertia of the rod and the angular velocity of the system. This additional term will affect the maximum angle of rotation, so it should be included in the equation.

Once we have the equation set up, we can plug in the given values for the mass of the rod, the velocity of the clay ball, and the length of the rod to solve for the maximum angle of rotation. It may also be helpful to draw a diagram and use trigonometric functions to relate the angle θ to the length of the rod and the position of the clay ball.

In conclusion, to determine the maximum angle of rotation of the rod with the attached clay ball, we need to consider the conservation of energy and the rotational motion of the system, and use the relevant equations to solve for the angle θ.