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A 1 kg ball is hung at the end of a rod 1 m long. If the system balances at a point on the rod one third of the distance from the end holding the mass, what is the mass of the rod?

Respuesta :

MathPhys

Answer:

2 kg

Explanation:

Assuming the rod's mass is uniformly distributed, the center of mass is at half the length.

Sum of the moments at the balance point:

-(Mg)(L/3) + (mg)(L/2 − L/3) = 0

(Mg)(L/3) = (mg)(L/2 − L/3)

(Mg)(L/3) = (mg)(L/6)

2M = m

M = 1 kg, so m = 2 kg.

The mass of the rod is 2 kg.

onyebuchinnaji

The mass of the rod on which balances the ball hung at one end is 2 kg.

The given parameters;

  • mass of the ball, m = 1 kg
  • length of the rod, L = 1 m

The one-third distance from the point holding the mass is calculated as follows;

p = ¹/₃ x 1 = 0.33 m

 

0                                 0.33 m                                                        1 m

------------------------------Δ----------------------------------------------------------

↓                                                                                                      ↓

W                                                                                                 1 kg

Take moment about the pivot;

W(0.33) = 1(1 - 0.33)

W(0.33) = 1(0.67)

[tex]W = \frac{0.67}{0.33} \\\\W = 2 \ kg[/tex]

Thus, the mass of the rod on which balances the ball hung at one end is 2 kg.

Learn more here:https://brainly.com/question/17152793

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