Question

A 750 g. 50-cm-long metal rod is free to rotate about a frictionless axle at one end. While at rest, the rod is given a short but sharp 1000 N hammer blow at the center of the rod, almed in a direction that causes the rod to rotate on the axde. The blow lasts a mere 1.8 ms. Part A What is the rod's angular velocity immediately after the blow? Express your answer to two significant figures and include the appropriate units Value Units Submit Beguest Answer

Answer 1

Final angular momentum = angular impulse given

I*w = mvr

0.750*0.5^2/3 *w = 1000*1.8e-3*0.50/2

w = [1000*1.8e-3*0.50/2]/(0.750*0.5^2/3)

= 7.2 rad/s answer

Answer 2

L(angular momentum) = Iw = T(torque)△t(time)

I of a rod about its end = (1/3)(m)(r)^2

Torque fomrula = rFsin(θ)

make sure you convert everything into meters, kg, and seconds

therefore (1/3)(m)(r^2)w = rFsin(θ)△t

with these values: (1/3)(.75)(.5^2)w = (.5)(1000)sin(90)(0.0018)/2

for some reason you divide by 2 on the right side, i'm not sure why .n.

when solving for w you should get w = (.45)/(.0625) = 7.2 rad/s