Question

A uniform stick 1.5 m long with a total mass of 300 g is pivoted at its center. A 3.0-g bullet is shot through the stick midway between the pivot and one end The bullet approaches at 250 m/s and leaves at 140 m/s.
With what angular speed is the stick spinning after the collision?
ω= rad/s

Answer 1

Answer:2.2 rad/s

Explanation:

Given

Length of stick=1.5 m

mass m=300 gm

mass of bullet $m_0=3 gm$

approach velocity of bullet=250 m/s

Final Velocity of Velocity=140 m/s

Distance of Pivot to bullet hit is $r=\frac{L}{4}=0.375 m$

Conserving Angular Momentum

Initial momentum $L_i=mv_i\cdot r$

$L_i=3\times 10^{-3}\times 250\times 0.375=0.281 kg.m^2/s$

$L_f=mv_f\cdot r$

$L_f=3\times 10^{-3}\times 140\times 0.375=0.157 kg.m^2/s$

Moment of inertia of rod about Pivot$=\frac{mL^2}{12}=0.05625 kg.m^2$

$L_{net}=I\times \omega _{stick}$

$0.281-0.157=0.05625\times \omega _{stick}$

$\omega _{stick}=\frac{0.124}{0.05625}=2.2 rad/s$

Answer 2

Answer

given,

length of stick = 1.5 m

mass of the stick = 300 g = 0.3 kg

mass of bullet, m = 3 g = 0.003 kg

initial velocity (v_1) = 250 m/s

final velocity (v_2) = 150 m/s

distance of the pivot point from center

    $r = \dfrac{H}{4}$

    $r = \dfrac{1.5}{4}$

           r = 0.375 m

angular momentum of stick after collision

$L_{stick} = L_i - L_f$

              = $mv_ir - mv_fr$

              = $mr(v_i-v_f)$

              = 0.003\times 0.375 \times (250-140)[/tex]

              = 0.12375 kg.m²/s

momentum of inertia

$I = \dfrac{1}{12}MH^2$

$I = \dfrac{1}{12}\times 0.3 \times 1.5^2$

angular speed of stick

$\omega_{stick}= \dfrac{L_{stick}}{I}$

                        = $\dfrac{0.12375}{0.05625}$

                        = 2.2 rad/s