Question

A uniform stick 1.5 m long with a total mass of 250 g is pivoted at its center. A 3.3-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?

Answer 1

Answer:

63.44 rad/s

Explanation:

mass of bullet = 3.3 g = 0.0033 kg

initial velocity of bullet $v_{1}$ = 250 m/s

final velocity of bullet $v_{2}$ = 140 m/s

loss of kinetic energy of the bullet = $\frac{1}{2}m(v^{2} _{1} - v^{2} _{2})$

==> $\frac{1}{2}*0.0033*(250^{2} - 140^{2} )$ = 70.785 J

this energy is given to the stick

The stick has mass = 250 g =0.25 kg

its kinetic energy = 70.785 J

from

KE = $\frac{1}{2} mv^{2}$

70.785 = $\frac{1}{2}*0.25*v^{2}$

566.28 = $v^{2}$

$v= \sqrt{566.28}$ = 23.79 m/s

the stick is 1.5 m long

this energy is impacted midway between the pivot and one end of the stick, which leaves it with a radius of 1.5/4 = 0.375 m

The angular speed will be

Ω = v/r = 23.79/0.375 = 63.44 rad/s