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A 34 kg bowling ball with a radius of 22 cm starts from rest at the top of an incline 4.4 m in height. Find the translational speed of the bowling ball after it has rolled to the bottom of the incline. (Assume that the ball is a uniform solid sphere.)The acceleration of gravity is 9.81 m/s 2 . Answer in units of m/s.

Respuesta :

Answer:

7.85 m/s.

Explanation:

Given,

Mass of the bowling ball, m = 34 Kg

radius, r = 0.22 cm

height of the inclination, h = 4.4 m

transnational velocity = ?

Moment of inertia of bowling ball,

[tex]I = \dfrac{2}{5}mr^2[/tex]

Using conservation of energy

[tex]mgh = \dfrac{1}{2}I\omega^2 + \dfrac{1}{2}mv^2[/tex]

We know that [tex]v = r\omega[/tex]

[tex]mgh = \dfrac{1}{2}( \dfrac{2}{5}mr^2)(\dfrac{v}{r})^2 + \dfrac{1}{2}mv^2[/tex]

[tex] m gh = 0.7 m v^2[/tex]

[tex]v =\sqrt{\dfrac{gh}{0.7}}[/tex]

[tex]v =\sqrt{\dfrac{9.81\times 4.4}{0.7}}[/tex]

[tex]v = 7.85\ m/s[/tex]

Speed of the bowling ball is equal to 7.85 m/s.

hamzaahmeds

This question involves the concepts of the law of conservation of energy, potential energy, and kinetic energy.

The translational speed of the ball at the bottom of the incline is "7.85 m/s".

According to the law of conservation of energy:

Loss in Potential Energy = Gain in Rotational Kinetic Energy + Gain in Translational Kinetic Energy

[tex]mgh =\frac{1}{2}{I}{\omega^2}+\frac{1}{2}mv^2[/tex]

where,

  • m = mass of ball = 34 kg
  • g = acceleration due to gravity = 9.81 m/s²
  • h = height = 4.4 m
  • I = moment of inertia of spherical ball = [tex]\frac{2}{5}mr^2[/tex]
  • v = translational speed = ?
  • ω = rotational speed = [tex]\frac{v}{r}[/tex]

Therefore,

[tex]mgh=\frac{1}{2}(\frac{2}{5}mr^2)(\frac{v}{r})^2+\frac{1}{2}mv^2\\\\gh = 0.2v^2+0.5v^2\\\\v=\sqrt{\frac{gh}{0.7}}\\\\v=\sqrt{\frac{(9.81\ m/s^2)(4.4\ m)}{0.7}}[/tex]

v = 7.85 m/s

Learn more about the law of conservation of energy here:

https://brainly.com/question/381281

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