Question

A pin-supported, vertically-oriented 1-m long thin rod is struck by a pellet at m down from the pin at the top. The mass of the pellet is and the mass of the rod is . The pellet has an initial velocty , and the thin rod is initially at rest. Matlab/Mathematica input: mp = 0.012; mr = 7; ry = 0.8 vpix = 260; vpiy = 350; What is the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod?

Answer 1

Answer:

the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s

Explanation:

Using the conservation of momentum of approach.

From the question; the pellet is hitting at a distance of 0.4 m down from the point of rotation of the rod.

So, the angular momentum of the system just before the collision occurs  with respect to the axis of the rotation is expressed by the formula:

$L_i ^ { ^ \to } = mp ( r_y } ^ { ^ \to } * v_{pi} ^ { ^ \to } )$    ----- equation (1)

The position vector can now be :

$x ^ { ^ \to }$ = $- 0.4 \ j \ m$

Also, given that :

$v_{p,i} ^ { ^ \to } = (280 \ i - 350 \ j) \ m/s$

Replacing the value into above equation (1); we have:

$L_i ^ { ^ \to } =0.012 ((- \ 0.4 \ j) *(280 \ i - 350 \ j ))$

$L_i ^ { ^ \to } =0.012 * 112 \ k$    (by using cross product )

$L_i ^ { ^ \to } = 1.344 k` \ \ kg m^2 s^{-1}$

However; the moment of inertia of the rod about the axis of rotation is :

$I_{rod} = \frac{1}{3}m_rl^2 \\ \\ I_{rod} = \frac{1}{3}*8*1^2 \\ \\ I_{rod} = \frac{8}{3} \ \ kg \ m^2$

Also, the moment of inertia of the pellet about the axis of rotation is:

$I_{pellet} = m_pr_y^2 \\ \\ I_{pellet} = 0.012 *0.4^2 \\ \\ I_{pellet} = 1.92*10^{-3} kg . m^2$

So, the moment of inertia of the rod +pellet system is:

$I = I_{rod}+I_{pellet}$

$I =( \frac{8}{3}+1.92 *10^{-3} )kg. m^2$

$I = 2.6686 \ kg. m^2$

The final angular momentum is :

$L_f ^ {^ \to} = I \omega { ^ {^ \to} } = 2.6686 \ \omega ^ {^ \to}$

The angular velocity of the rod $\omega$ is determined by equating the angular momentum just before the collision with the final angular momentum (i.e after the collision).

So;

$L_f ^ {^ \to} = L_i ^ { ^ \to}$

$2.6686 \omega ^ {^ \to} = 1.344 \ k ^ {^ \to}$

$\omega ^ {^ \to} = \frac{1.344 \ k`}{2.6686}$

= 0.5036 k` rad/s

Hence; the angular velocity of the rod immediately after being struck by the pellet, provided that the pellet gets lodged in the rod is = 0.5036 k` rad/s