A uniform rod of mass 100 g and length 50.0 cm rotates in a horizontal plane abo

Question

A uniform rod of mass 100 g and length 50.0 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small beads, each of mass 31.0 g, are mounted on the rod so that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10.0 cm on each side of center: at which time the system rotates at an angular speed of 18.0 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. (a) Find the angular speed of the system at the instant the beads reach the ends of the rod. rad/s(b) What if the beads fly off the ends? What is the angular speed of the rod after this occurs? remains constant varies linearly as a function of time varies exponentially as a function of time becomes zero

Explanation / Answer

a)Mass of rod=M=100 gm = 0.1 kg

Length of rod =L= 50 cm=0.50 m

Mass of beads = m =31 gm = 0.031 kg

Intial distance of beads from rod center=d=10 cm = 0.1 m

Initial mass moment of inertia of system = (ML²)/12 + (2*m*d²) =(0.1*0.5²/12) +(2*0.031*0.1²)=0.002703 kg-m²

Initial angular velocity =wi18 ra/s

Final distance of beads from center of rod =0.25m

Final mass moment of inertia of the system =

(0.1*0.5²/12) +(2*0.031*0.25²)=0.005958 kg-m²

Let final angular velocity be wf

From conservation of angular momentum

Initial angular momentum = final angular momentum

0.002703*18=0.005958*wf

wf = 8.166 rad/s

b)answer is remains constant. This is because even if the beads fly off, momentum of the system remains the same.

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

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