A uniform 1.8 m long rod with a mass of 1.17 kg is suspended vertically from a f

Question

A uniform 1.8 m long rod with a mass of 1.17 kg is suspended vertically from a frictionless pin at its upper end. A 4 gram bullet with a speed of 238 m/s strikes the rod 1.3 m below the pivot point.

Determine how high the bottom of the rod swings above the datum if the bullet remains embedded in the rod.

Determine how high the bottom of the rod swings above the datum if the bullet goes through the rod and leaves with a speed of 106 m/s.

Hint #1: Use conservation of angular momentum to analyze the collision. Hint #2: You can neglect the small change in the center of mass location due to the bullet now being in the rod, but you cannot ignore the change it causes for the mass moment of inertia. Hint #3: Use COE (during the swing) to find the height change for the center of mass. Hint #4: Use ratios to relate the change in height of the center of mass to the change in height of the end of the rod.

Explanation / Answer

Here ,

let the angular speed of the rod after collision is w

Using conservation of angular momentum

I * w = m * v * r

(1.17 * 1.8^2/3 + 0.004 * 1.3^2) * w = 0.004 * 1.3 * 238

solving for w

w = 0.974 rad/s

let the height increased is h

0.5 * (1.17 * 1.8^2/3 + 0.004 * 1.3^2) * 0.974^2 = (1.17 + 0.004) * 9.8 * h

solving for h

h = 0.0523 m

the end of the rod is 0.0523 m

------------------------

for the final speed of bullet , vf = 106 m/s

Using conservation of angular momentum

I * w = m * (vf - vi) * r

(1.17 * 1.8^2/3 + 0.004 * 1.3^2) * w = 0.004 * 1.3 * (238 - 106)

solving for w

w = 0.5403 rad/s

let the height increased is h

0.5 * (1.17 * 1.8^2/3 + 0.004 * 1.3^2) * 0.5403^2 = (1.17 + 0.004) * 9.8 * h

solving for h

h = 0.016 m

the end of the rod is 0.016 m

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.