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A uniform stick 1.3 m long with a total mass of 260 g is pivoted at its center. A 3.6 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?
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by conservation of energy ,
kinetic energy lost by the bullet = rotational kinetic energy gained by stick. hence, 1/2 m1 (v1^2 - v2^2 ) = 1/2 I w^2 = 1/2 m2 l^2/12 x w^2,
where m1 is mass of bullet, m2 is mass of stick and 'w' is angular velocity of the stick after collision.
ie., w^2 = 12 m1 (v1^2 - v2^2 ) / m2 l^2
= 4217.75 ( after using the values and simplifying)
and hence, w = 64.944 = 65 rad/sec
This answer is wrong!?
A uniform stick 1.3 m long with a total mass of 260 g is pivoted at its center. A 3.6 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?
~~~~~~~~~~~~
by conservation of energy ,
kinetic energy lost by the bullet = rotational kinetic energy gained by stick. hence, 1/2 m1 (v1^2 - v2^2 ) = 1/2 I w^2 = 1/2 m2 l^2/12 x w^2,
where m1 is mass of bullet, m2 is mass of stick and 'w' is angular velocity of the stick after collision.
ie., w^2 = 12 m1 (v1^2 - v2^2 ) / m2 l^2
= 4217.75 ( after using the values and simplifying)
and hence, w = 64.944 = 65 rad/sec
This answer is wrong!?