Respuesta :
Answer:
Height reached by the ball, h = 3.57 meters
Explanation:
It is given that,
Mass of the disk, m = 42 kg
Diameter of the disk, d = 3.2 m
Radius, r = 1.6 m
Angular speed of the disk, [tex]\omega=4.27\ rad/s[/tex]
The kinetic energy of the disk is equal to its potential energy. Using the conservation of energy as :
[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2=mgh[/tex]
[tex]\dfrac{1}{2}mv^2+\dfrac{1}{2}(mr^2/2)\omega^2=mgh[/tex]
[tex]\dfrac{1}{2}v^2+\dfrac{1}{2}(r^2/2)\omega^2=gh[/tex]
[tex]\dfrac{1}{2}(r\omega)^2+\dfrac{1}{2}(r^2/2)\omega^2=gh[/tex]
[tex]\dfrac{1}{2}(1.6\times 4.27)^2+\dfrac{1}{2}(1.6^2/2)\times 4.27^2=gh[/tex]
[tex]h=\dfrac{35.0071}{9.8}[/tex]
h = 3.57 meters
So, the solid disk will reach to a height of 3.57 meters.
The height attained by the disk on the hill is 3.57 m.
The given parameters;
- mass of the disk, m = 42 kg
- diameter of the disk, d = 3.2 m
- angular speed of the disk, ω = 4.27 rad/s
The height attained by the disk is calculated by applying the law of conservation of energy as follows;
[tex]\frac{1}{2}mv^2 + \frac{1}{2}I \omega ^2 = mgh\\\\mv^2 + I \omega ^2 = 2mgh\\\\mv^2 + (\frac{mr^2}{2} ) \omega ^2 = 2mgh\\\\v^2 + \frac{1}{2} r^2 \omega ^2 = 2gh\\\\(\omega r)^2 + 0.5(\omega r )^2 = 2gh\\\\1.5 (\omega r)^2 = 2gh\\\\h = \frac{1.5 (\omega r)^2}{2g} \\\\h = \frac{1.5 \times (4.27 \times 1.6)^2 }{2\times 9.8} \\\\h = 3.57 \ m[/tex]
Thus, the height attained by the disk on the hill is 3.57 m.
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