Answer:
9.6 rad/s
Explanation:
[tex]L[/tex] = length of the metal rod = 50 cm = 0.50 m
[tex]M[/tex] = Mass of the long metal rod = 780 g = 0.780 kg
Moment of inertia of the rod about one end is given as
[tex]I = \frac{ML^{2}}{3} = \frac{(0.780)(0.50)^{2}}{3} = 0.065 kgm^{2}[/tex]
[tex]F[/tex] = force applied by the hammer blow = 1000 N
Torque produced due to the hammer blow is given as
[tex]\tau = \frac{FL}{2}[/tex]
[tex]\tau = \frac{(1000)(0.50)}{2}[/tex]
[tex]\tau = 250 Nm[/tex]
[tex]t[/tex] = time of blow = 2.5 ms = 0.0025 s
[tex]w[/tex] = Angular velocity after the blow
Using Impulse-change in angular momentum, we have
[tex]I w = \tau t\\(0.065) w = (250) (0.0025)\\w = 9.6 rads^{-1}[/tex]
