A skater of mass 40 kg is carrying a box of mass 5 kg. The skater has a speed of 5 m/s with respect to the floor and is gliding
1 answer:
For the part a) we need only the momentum of the box and we have the data to find it.
Momentum is given by,
where clearly, p is the momentum, m the mass of the box and v is the velocity.
Substituting,
For part b) we need an analysis of the situation. We understand that the box on a surface that has no friction will continue to rotate at the same speed previously defined. The box can only stop with friction, so,
<em>It is the same that part a)</em>
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The formula for the rotational kinetic energy is
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,
where I is the moment of inertia. This is just mass times the square of the perpendicular distance to the axis of rotation. In other words, the radius of the propeller or this is equivalent to the length of the rod. ω is the angular velocity. We determine I and ω first.
ω = 573 rev/min * (2π rad/rev) * (1 min/60 s) = 60 rad/s
Then,