Respuesta :
Answer:
[tex]\dfrac{K_t}{K_r}=\dfrac{5}{2}[/tex]
Explanation:
Given that,
Mass of the bowling ball, m = 5 kg
Radius of the ball, r = 11 cm = 0.11 m
Angular velocity with which the ball rolls, [tex]\omega=2.8\ rad/s[/tex]
To find,
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball.
Solution,
The translational kinetic energy of the ball is :
[tex]K_t=\dfrac{1}{2}mv^2[/tex]
[tex]K_t=\dfrac{1}{2}m(r\omega)^2[/tex]
[tex]K_t=\dfrac{1}{2}\times 5\times (0.11\times 2.8)^2[/tex]
The rotational kinetic energy of the ball is :
[tex]K_r=\dfrac{1}{2}I \omega^2[/tex]
[tex]K_r=\dfrac{1}{2}\times \dfrac{2}{5}mr^2\times \omega^2[/tex]
[tex]K_r=\dfrac{1}{2}\times \dfrac{2}{5}\times 5\times (0.11)^2\times (2.8)^2[/tex]
Ratio of translational to the rotational kinetic energy as :
[tex]\dfrac{K_t}{K_r}=\dfrac{5}{2}[/tex]
So, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is 5:2
The ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is [tex]\frac{5}{2}[/tex] or 2.50.
Given the data in question;
- Mass of the bowling ball; [tex]m = 5.0kg[/tex]
- Radius; [tex]r = 11.0cm = 0.11m[/tex]
- Angular velocity; [tex]w = 2.80rad/s[/tex]
Ratio of [tex]E_{translational}[/tex] to the [tex]E_{rotational }[/tex]; [tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}} = \ ?[/tex]
Translational and Rotational kinetic Energy
Rotational energy or angular kinetic energy is simply the kinetic energy due to the rotation of a rigid body while translational kinetic energy is the same as one-half the product of mass and the square of the velocity of the body.
They are expressed as;
[tex]E_{rotational} = \frac{1}{2}Iw^2[/tex]
Where;
- [tex]w[/tex] is the angular speed or velocity.
- [tex]I[/tex] is the moment of inertia of the bowling ball ( [tex]I = \frac{2}{5}mR^2[/tex] ).
Hence, [tex]E_{rotational} = \frac{1}{2}(\frac{2}{5}mR^2 )w^2[/tex]
[tex]E_{translational} = \frac{1}{2}mv^2[/tex]
Where;
- m is mass of the object
- v is the linear velocity ( [tex]v = R*w[/tex] )
Hence, [tex]E_{translational} = \frac{1}{2}m(R*w)^2[/tex]
To determine the translational kinetic energy of the bowling ball, we substitute our given values into the equation above.
[tex]E_{translational} = \frac{1}{2}m( R*w)^2\\ \\E_{translational} = \frac{1}{2} * 5.0kg * 0.0121m^2 * 7.84rad/s^2[/tex]
To determine the rotational kinetic energy of the bowling ball, we substitute our given values into the equation above.
[tex]E_{rotational} = \frac{1}{2}( \frac{2}{5}mR^2)w^2\\ \\E_{rotational} = \frac{1}{2}( \frac{2}{5} * 5.0kg* (0.11m)^2)(2.80rad/s)^2\\ \\E_{rotational} = \frac{2}{10} * 5.0kg * 0.0121m^2 * 7.84rad/s^2[/tex]
So, Ratio of [tex]E_{translational}[/tex] to the [tex]E_{rotational }[/tex]; [tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}}[/tex]
[tex]R_{\frac{E_{translational}}{E_{rotational}}}= \frac{E_{translational}}{E_{rotational}}\\\\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1}{2} * 5.0kg * 0.0121m^2 * 7.84rad/s^2}{\frac{2}{10} * 5.0kg * 0.0121m^2 * 7.84rad/s^2}\\\\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1}{2} }{\frac{2}{10} }\\ \\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{\frac{1*10}{2} }{2}\\ \\R_{\frac{E_{translational}}{E_{rotational}}} = \frac{5}{2} = 2.50[/tex]
Therefore, the ratio of the translational kinetic energy to the rotational kinetic energy of the bowling ball is [tex]\frac{5}{2}[/tex] or 2.50.
Learn more about translational kinetic energy and rotational kinetic energy: https://brainly.com/question/13926915
