Question

A uniform 140 g rod with length 57 cm rotates in a horizontal plane about a fixed, vertical, frictionless pin through its center. Two small 30 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 11 cm on each sides of the center, at which time the system rotates at an angular speed of 23 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 correct answer is "12 rad/s"

Explanation:

The given values are,

Mass of rod,

M = 140 g

i.e.,

   = 0.14 kg

Length,

L = 57 cm

i.e.,

  = 0.57 m

Mass of beads,

M = 30 g

i.e.,

   = 0.03 kg

Angular speed,

r = 11 cm

i.e.,

 = 0.11 m

Now,

The inertia of rods will be:

=  $\frac{1}{12}ML ^2$

On substituting the values, we get

=  $\frac{1}{12}\times 0.14\times (0.57)^2$

=  $0.0037905 \ kg-m^2$

The inertia of beads will be:

=  $mr^2$

On substituting the values, we get

=  $0.03\times (0.11)^2$

=  $0.000726 \ kg-m^2$

The total inertia will be:

=  $Inertia \ of \ rods+Inertia \ of \ beads$

=  $0.0037905 + 0.000726$

=  $0.0045165 \ kg-m^2$

now,

The angular speed of the system will be:

⇒ $L_1w_1=L_2w_2$

On substituting the values in the above equation, we get

⇒ $0.0045165\times 23 = (0.0037905 + (0.03\times 0.285^2)\times 2 )\times w_2$

⇒ $0.1038795 = 0.0037905 + (0.00243675\times 2 )\times w_2$

⇒             $w_2 = 12 \ rad/s$