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A small ball is attached to one end of a spring that has an unstrained length of 0.201 m. The spring is held by the other end, and the ball is whirled around in a horizontal circle at a speed of 3.41 m/s. The spring remains nearly parallel to the ground during the motion and is observed to stretch by 0.0176 m.

Required:
By how much would the spring stretch if it were attached to the ceiling and the ball allowed to hang straight down, motionless?

Respuesta :

onyebuchinnaji

The extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.

The given parameters;

  • unstrained length, l₁ = 0.201 m
  • extension of the string, x = 0.0176 m
  • speed of the ball, v = 3.41 m/s

The radius of the circular path when spring is stretched is calculated as;

R = l₁ + x

R = 0.201 + 0.0176

R = 0.2186 m

The spring constant is calculated as follows;

[tex]F = ma\\\\Kx = \frac{mv^2}{R} \\\\K = \frac{mv^2}{Rx} \\\\K = \frac{(3.41)^2 m}{0.2186 \times 0.0176} \\\\K = 3,022.4 m \ N/m[/tex]

The extension of the spring when the ball is allowed to hang straight down, motionless;

[tex]F = mg\\\\Kx=mg\\\\x = \frac{mg}{K} \\\\x = \frac{9.8 m}{3022.4 m} \\\\x = 0.0032 \ m[/tex]

Thus, the extension of the spring when the ball is allowed to hang straight down, motionless is 0.0032 m.

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