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Tresset [83]
2 years ago
15

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?
Physics
1 answer:
Andru [333]2 years ago
5 0

Answer:

<em>63.44 rad/s</em>

<em></em>

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 =<em> 63.44 rad/s</em>

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