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SIZIF [17.4K]
3 years ago
13

skater spins over a point at a speed of 3.0 rotations per second then the momentum of inertia is 0.60 kg.M2, what is its angular

momentum?
Physics
1 answer:
laiz [17]3 years ago
8 0

Answer:

L=11.3\ kg-m^2/s

Explanation:

Given that,

Angular speed of a skater, \omega=3\ rot/s=18.84\ rad/s

The moment of inertia of the skater, I = 0.6 kg-m²

We need to find the angular momentum of the skater. The formula for the angular momentum of the skater is given by :

L=I\omega

Substitute all the values,

L=0.6\times 18.84\\\\L=11.3\ kg-m^2/s

So, its angular momentum is equal to 11.3\ kg-m^2/s.

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hello your question is incomplete hence I will give you a general answer on how a stunt cyclist performs his stunts successfully

answer ;

The stunt cyclist trying to perform some sort of stunt with the bicycle will be faced with some forces like the static fiction between the tires of the bicycle and wall and also a centripetal force.

In order to overcome the centripetal force the minimal grip of the bicycle must be equal to the weight and the speed at which the cyclist must go can be represented as :

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Explanation:

The stunt cyclist trying to perform some sort of stunt with the bicycle will be faced with some forces like the static fiction between the tires of the bicycle and wall and also a centripetal force.

In order to overcome the centripetal force the minimal grip of the bicycle must be equal to the weight and the speed at which the cyclist must go can be represented as :

V_{min} = \sqrt{\frac{r.g}{u} }

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