Question

A rotating table completes one rotation in 10sec. and it's moment of inertia is 100kg-m2. A person of mass 50kg stands at the centre of rotating table. If the person moves 2meter from the center, the angular velocity of rotating table is?

Answer 1

The angular velocity of rotating table is 0.2095$(\frac{rad}{s})$, it will complete one rotation in 30 second.

Explanation:

1. Here no external torque is acting so angular momentum of table is constant.

  Means

Initial angular momentum= final angular momentum

$I_{1}\omega _{1}=I_{2}\omega _{2}$      ...1)

Where

$I_{1}$ = initial mass moment of inertia = 100 $kgm^{2}$

$I_{2}$ = Final mass moment of inertia = $100 + MX^{2}=100 + 50\times 2^{2}=300 (kgm^{2})$        ...2)

2.  In equation 2)

$MX^{2}$= mass moment of inertia of man with respect to centre of rotation

              M = mass of man =50 kg

              X = distance from centre of rotation = 2 m

3. In equation 1)

   $\omega _{1}$ = initial angular velocity = $\frac{2\pi }{10} (\frac{rad}{s})$  

4. Now from equation 1)

     $I_{1}\omega _{1}=I_{2}\omega _{2}$  

     $100\times (\frac{2\pi }{10})=300\times \omega _{2}$   ...3)

5. On solving equation 3)

   We get

  $\omega _{2}= \frac{2\pi }{30}= \frac{2\times 22}{30\times 7}=0.2095 (\frac{rad}{s})$

  It  angular speed of table is 0.2095 radiant per second we can also say table complete one rotation in 30 second.

   

           

Answer 2

Answer:

π/15

Explanation:

When person moves 2m then moment of inertia will be sum of MOI

I'=I+mr²

=100+50×2²

=100+200

=300

So, according to conservation of angular momentum the new angular velocity become 1/3 of the initial because the moment of inertia gets 3 times

=10×3=30

Therefore: w=2π/30

π/15

Hope it helps you..all the best