Question

A boy of mass 30kg is running with a velocity 3 m/s on the ground just tangential to the merry go round which is at rest.it had radius 2m,mass 120kgants radius of gyration is 1m.if the boy suddenly jumps into the merry go round , the angular velocity aqquired by the system is

Answer 1

Given:

A boy of mass 30kg is running with a velocity 3 m/s on the ground just tangential to the merry go round which is at rest.it had radius 2m,mass 120kgants radius of gyration is 1m.

To find:

Angular velocity acquired by the system when boy jumps on the merry-go round.

Calculation:

In this type of questions , we need to apply the concept of conservation of angular momentum since no is External Torque is involved.

Let mass of boy be m and mass of merry-go-round be M ; k denotes radius of gyration.

$\rm{ \therefore \: L = constant}$

$\rm{ = > \:( I1)( \omega1)= (I1 + I2) \omega}$

$\rm{ = > \:(m {r}^{2} )( \dfrac{v1}{r} )= (I1 + I2) \omega}$

$\rm{ = > \:(m {r}^{2} )( \dfrac{v1}{r} )= ( M{r}^{2} + M {k}^{2} ) \omega}$

$\rm{ = > \:(mr) (v1)= (M {r}^{2} + M {k}^{2} ) \omega}$

Putting available values in SI unit:

$\rm{ = > \:30 \times 2 \times 3= M ({r}^{2} + {k}^{2} ) \omega}$

$\rm{ = > \:180= M ({r}^{2} + {k}^{2} ) \omega}$

$\rm{ = > \:180= M \{{(2)}^{2} + {(1)}^{2} \}\omega}$

$\rm{ = > \:180= M \{5 \}\omega}$

$\rm{ = > \:180= 120 \times5\omega}$

$\rm{ = > \:3= 2 \times5\omega}$

$\rm{ = > \:3= 10\omega}$

$\rm{ = > \:\omega = 0.3 \: rad {s}^{ - 1} }$

So, final answer is:

$\boxed{\bf{\omega = 0.3 \: rad {s}^{ - 1} }}$