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A barbell spins around a pivot at its center at A. The barbell consists of two small balls, each with mass 450 grams (0.45 kg), at the ends of a very low mass rod of length d = 20 cm (0.2 m; the radius of rotation is 0.1 m). The barbell spins clockwise with angular speed 120 radians/s.

What is the speed of ball 1?

Respuesta :

Answer:

The speed of ball is 12 [tex]\frac{m}{s}[/tex]

Explanation:

Given:

Mass of ball [tex]m = 0.45[/tex] kg

Radius of rotation [tex]r = 0.1[/tex] m

Angular speed [tex]\omega = 120 \frac{rad}{s}[/tex]

Here barbell spins around a pivot at its center and barbell consists of two small balls,

From the formula of speed in terms of angular speed,

  [tex]v = r \omega[/tex]

Where [tex]v =[/tex] speed of ball

  [tex]v = 120 \times 0.1[/tex]

  [tex]v = 12 \frac{m}{s}[/tex]

Therefore, the speed of ball is 12 [tex]\frac{m}{s}[/tex]

onyebuchinnaji

The linear speed of the ball for the circular motion is determined as 12 m/s.

The given parameters;

  • mass of each ball, m = 450 g = 0.45 kg
  • length of the rod, L = 0.2 m
  • radius of the rod, r = 0.1 m
  • angular speed of the ball, ω = 120 rad/s

The linear speed of the ball is calculated as follows;

v = ωr

where;

  • ω is the angular speed of the ball
  • r is the radius of circular motion of the ball

The linear speed of the ball is calculated as follows;

v = ωr

v = 120 x 0.1

v = 12 m/s

Thus, the linear speed of the ball for the circular motion is determined as 12 m/s.

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