A block of mass 0.1 kg is attached to a cord passing through a hole in a horizontal frictionless horizontal surface as shown in the figure. The block is originally revolving at a distance of 0.4 m from the hole, with an angular velocity of 2 rad s- 1 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.2 m . Considering the block to be point mass, calculate the new angular velocity (in rad s- 1 ).

Question

A block of mass 0.1 kg is attached to a cord passing through a hole in a horizontal frictionless horizontal surface as shown in the figure. The block is originally revolving at a distance of 0.4 m from the hole, with an angular velocity of 2 rad s- 1 . The cord is then pulled from below, shortening the radius of the circle in which the block revolves to 0.2 m . Considering the block to be point mass, calculate the new angular velocity (in rad s- 1 ). <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-qkwo7ziclnoj.png" /> (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

By conservation of angular momentum I1ω 1=I2ω 2 ⇒ MR12ω 1=MR22ω 2 ⇒ ω2=((R1/R2))2 ω1 =((0 .4/0 .2))2× 2 =8rad s- 1

By conservation of angular momentum I1​ω1​=I2​ω2​ ⇒MR12​ω1​=MR22​ω2​ ⇒ω2​=(R2​R1​​)2ω1​ =(0.20.4​)2×2 =8rads−1

By conservation of angular momentum
$I_{1} ω_{1} = I_{2} ω_{2}$
$\Rightarrow M R_{1}^{2} ω_{1} = M R_{2}^{2} ω_{2}$
$\Rightarrow ω_{2} = (\frac{R _{1}}{R _{2}})^{2} ω_{1}$
$= (\frac{0.4}{0.2})^{2} \times 2$
$= 8 r a d s^{- 1}$