Long uniform rod forumla: (1/12)*(M)*(L^2) A uniform stick 1.5m long with a tota

Question

Long uniform rod forumla: (1/12)*(M)*(L^2)

A uniform stick 1.5m long with a total mass of 270g 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/sand leaves at 140 m/s. (Figure 1)

Part A

With what angular speed is the stick spinning after the collision?

Express your answer using two significant figures.

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Figure 1 of 1

A uniform stick 1.5m long with a total mass of 270g 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/sand leaves at 140 m/s. (Figure 1)

Part A

With what angular speed is the stick spinning after the collision?

Express your answer using two significant figures.

SubmitMy AnswersGive Up

Figure 1 of 1

Explanation / Answer

initial angular momentum of the bullet = m*v*R

= 3.6*10^-3*250*1.5/4

= 0.3375 kg.m^2/s

intital angular momentum of the stick = 0

total initial angualr momentum, L1 = 0.3375 kg.m^2/s

final angular momentum of the bullet = m*v*R

= 3.6*10^-3*140*1.5/4

= 0.189 kg.m^2/s

let w is the final angular velocity

final angular momentum of the stick = I*w

= (m*L^2/12)*w

= (0.27*1.5^2/4)*w

= 0.050625*w

total final angualr momentum, L2 = 0.189 + 0.050625*w kg.m^2/s

but, L1 = L2

0.3375 = 0.189 + 0.050625*w

==>   w = (0.3375 - 0.189)/0.050625
   = 2.933 rad/s <<<<<<Answer

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