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Liula [17]
3 years ago
15

A ball is tied to a 1.3 m string and whirled in a circular motion with a force of 220 N. If its speed is 6 m/s, (a) what is it c

entripetal acceleration? (b) What is the mass of the ball?
Physics
1 answer:
Studentka2010 [4]3 years ago
6 0

The centripetal acceleration is 27.692 kg m / s^2.

The mass of the ball is 7.94 kg.

<u>Explanation:</u>

  • The curved path of an object experience a force known as centripetal force. Its direction is always pointing to the motion of the body towards the center of curvature of the path.

            centripetal force Fc = (mass * square of velocity) / radius

where m represents mass in kg

          v represents  velocity

          r represents radius in a meter.

  • The acceleration experienced in uniform circular motion is called as centripetal acceleration.It always points toward the center of rotation and is perpendicular to the linear velocity.
  •                     centripetal acceleration = v^2 / r

                                                             = (6 * 6) / 1.3

                                                             = 27.692 kg m / s^2.

 The mass of the ball is found by using the centripetal force formula,

  •                               centripetal force Fc = m v^2 / r

                                                      m = (F * r) / v^2

                                                          =  (220 * 1.3) / 36

                           mass of the ball  m = 7.94 kg.

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Answer:

Part a)

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Part b)

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Part c)

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Part d)

v = \frac{1}{2}\sqrt{13Rg}

Part e)

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Part g)

X = \frac{13R}{4}

Explanation:

Initial speed of the launch is given as

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angle = \theta_i degree

Now the two components of the velocity

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similarly we have

v_y = v_i sin\theta_i

Part a)

Now we know that horizontal range is given as

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Part b)

Now the speed of the ball in x direction is always constant

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R = v_x T

R = v_x 2\sqrt{\frac{R}{3g}}

v_x = \frac{\sqrt{3Rg}}{2}

Part c)

Initial vertical velocity is given as

v_y = v_i sin\theta_i

v_i sin\theta = \sqrt{Rg/3}

Part d)

Initial speed is given as

v = \sqrt{v_x^2 + v_y^2}

so we will have

v = \sqrt{Rg/3 + 3Rg/4}

v = \frac{1}{2}\sqrt{13Rg}

Part e)

Angle of projection is given as

tan\theta_i = \frac{v_y}{v_x}

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\theta_i = 33.7 degree

Part f)

If we throw at same speed so that it reach maximum height

then the height will be given as

H = \frac{v^2}{2g}

H = \frac{13R}{8}

Part g)

For maximum range the angle should be 45 degree

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X = \frac{v^2}{g}

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