Is a man kicking ball potential or kinetic?
1 answer:
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Answer:
M₂ = M then L₂ = L
M₂> M then L₂ = \frac{M}{M_{2}} L
Explanation:
This is a static equilibrium exercise, to solve it we must fix a reference system at the turning point, generally in the center of the rod. By convention counterclockwise turns are considered positive
∑ τ = 0
The mass of the rock is M and placed at a distance, L the mass of the rod M₁, is considered to be placed in its center of mass, which by uniform e is in its geometric center (x = 0) and the triangular mass M₂, with a distance L₂
The triangular shape of the second object determines that its mass can be considered concentrated in its geometric center (median) that tapers with a vertical line if the triangle is equilateral, the most used shape in measurements.
M L + M₁ 0 - m₂ L₂ = 0
M L - m₂ L₂ = 0
L₂ = L
From this answer we have several possibilities
* if the two masses are equal then L₂ = L
* If the masses are different, with M₂> M then L₂ = \frac{M}{M_{2}} L
Answer:
Explanation:
We can solve this problem using conservation of angular momentum.
The angular momentum is
where is the position and
the linear momentum.
We also know that the torque is
but, as the linear momentum is this means that is parallel to the velocity, and the first term must equal zero
so
But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so
this means, for the angular momentum measure from the rod:
that means :
So, the magnitude of initial angular momentum is :
but the angle is 90°, so:
We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:
For our final angular momentum we have:
and the radius is 0.250 m and the mass is 2.00 kg
but, as the angular momentum is constant, this must be equal to the initial angular momentum