Question

A uniform 190 g rod with length 43 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 38 g beads are mounted on the rod such that they are able to slide without friction along its length. Initially the beads are held by catches at positions 10 cm on each sides of the center, at which time the system rotates at an angular speed of 12 rad/s. Suddenly, the catches are released and the small beads slide outward along the rod. Find the angular speed of the system at the instant the beads reach the ends of the rod. Answer in units of rad/s.

Answer 1

Answer:

The angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s

Explanation:

Moment of inertia is given as;

I = ¹/₁₂×ML² + 2mr²

where;

I is the moment of inertia

M is the mass of the rod = 0.19 kg

L is the length of the rod = 0.43 m

m is the mass of the bead = 0.038 kg

r is the distance of one bead

Initial moment of inertial is given as;

$I_i = \frac{1}{12}ML^2 +2mr_1^2$

Final moment of inertia is also given as

$I_f= \frac{1}{12}ML^2 +2mr_2^2$

Angular momentum is the product of angular speed and moment of inertia;

= Iω

From the principle of conservation of angular momentum;

$I_i \omega_i = I__f } \omega_f$

$(\frac{1}{12}ML^2 +2mr_1^2) \omega_i = (\frac{1}{12}ML^2 +2mr_2^2) \omega_f$

Given;

ωi = 12 rad/s

r₁ = 10.0 cm = 0.1 m

r₂ = 10.0cm/4 = 2.5 cm = 0.025 m

Substitute these values in the above equation, we will have;

$(\frac{1}{12}*0.19*(0.43)^2 +2*0.038(0.1)^2) 12 = (\frac{1}{12}*0.19*(0.43)^2 +2*0.038*(0.025)^2) \omega_f\\\\0.04425 =0.002975\ \omega_f\\\\\omega_f = \frac{0.04425}{0.002975} = 14.87\ rad/s$

Therefore, the angular speed of the system at the instant the beads reach the ends of the rod is 14.87 rad/s