Un electrón en un tubo de rayos catódicos acelera desde el reposo con una aceleración constante de 5.33x10¹²m/s² durante 0.150μs
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Answer:
The torque on the child is now the same, τ.
Explanation:
- It can be showed that the external torque applied by a net force on a rigid body, is equal to the product of the moment of inertia of the body with respect to the axis of rotation, times the angular acceleration.
- In this case, as the movement of the child doesn't create an external torque, the torque must remain the same.
- The moment of inertia is the sum of the moment of inertia of the merry-go-round (the same that for a solid disk) plus the product of the mass of the child times the square of the distance to the center.
- When the child is standing at the edge of the merry-go-round, the moment of inertia is as follows:
- So, τ = 3/2*m*r²*α (2)
- When the child moves to a position half way between the center and the edge of the merry-go-round, the moment of inertia of the child decreases, as the distance to the center is less than before, as follows:
- Since the angular acceleration increases from α to 2*α, we can write the torque expression as follows:
τ = 3/4*m*r² * (2α) = 3/2*m*r²
same result than in (2), so the torque remains the same.