Respuesta :
Answer
According the conservation of energy
[tex]\dfrac{1}{2}mv_i^2+\dfrac{1}{2}I\omega_i^2+0 = mgh + 0[/tex]
I for ball = [tex]\dfrac{2}{3}mr^2[/tex]
[tex]\dfrac{1}{2}mv_i^2+\dfrac{1}{2} \dfrac{2}{3}mr^2\omega_i^2= mgh[/tex]
[tex]\omega_i = \dfrac{v_i}{r}[/tex]
[tex]v_i^2+\dfrac{2}{3}r^2(\dfrac{v_i}{r})^2 = 2gh[/tex]
[tex]v_i^2+\dfrac{2}{3}v_i^2 = 2gh[/tex]
[tex]v_i^2+[1+\dfrac{2}{3}]=2gh[/tex]
[tex]v_i^2\dfrac{5}{3}=2gh[/tex]
[tex]v_i=\sqrt{\dfrac{6gh}{5}}=\sqrt{\dfrac{6\times 9.8\times 5}{5}}[/tex]
[tex]v_i = 7.67\ m/s[/tex]
a) [tex]\omega_i = \dfrac{v_i}{R}[/tex]
[tex]\omega_i = \dfrac{7.67}{0.113}[/tex]
[tex]\omega_i =67.87\ rad/s[/tex]
b) [tex]K_{rot} = \dfrac{1}{2}\dfrac{2}{3}mR^2\omega_i^2[/tex]
[tex]K_{rot} = \dfrac{1}{3}\times 0.426\times 0.112^2\times 67.87^2[/tex]
[tex]K_{rot} = 8.205\ J[/tex]
The rate at which the soccer ball was rotating at the base of the hill is 67.86 rad/s and the rotational kinetic energy is 8.35 J.
What is rotational kinetic energy?
Rotational kinetic energy is a type of energy, which a body is gains due to the rotational motion. The rotational kinetic energy of a body can be found with the following formula.
[tex]KE_{rotational}=\dfrac{1}{2}I\omega^2[/tex]
Here, (I) is the moment of inertia around the axis of rotation, and (ω) is the angular speed of the body.
For the rotational motion, the law of conservation of energy can be given as,
[tex]\dfrac{2}{6}mr^2\omega ^2+\dfrac{1}{2}mr^2\omega ^2=mgh\\\dfrac{2}{6}r^2\omega ^2+\dfrac{1}{2}r^2\omega ^2=gh\\\dfrac{5}{6}r^2\omega ^2=gh[/tex]
Here, (r) is the radius, (ω) is the angular velocity, (g) is the gravitational energy and (h) height.
- (a) The rate at which it was rotating at the base of the hill
The soccer ball diameter is 22.6 cm. Thus, the radius of it is,
[tex]r=\dfrac{22.6}{2}\times{10^{-2}}\rm\; m\\r=11.6\times{10^{-2}}\rm\; m[/tex]
It reaches to a maximum height of 5.00 m above the base of the hill. Plug the values in the above expression to find the angular velocity as,
[tex]\dfrac{5}{6}(11.6\times10)^{-2}^2\omega ^2=(9.8)(5)\\\omega=67.86\rm \; rad/s[/tex]
- (b) The rotational kinetic energy
The mass of the soccer ball is 426 g or 0.426 kg. By the above formula,
[tex]KE_{rotational}=\dfrac{1}{2}I\omega^2\\KE_{rotational}=\dfrac{1}{2}\times\dfrac{2}{3}mr^2\omega^2\\KE_{rotational}=\dfrac{1}{2}\times\dfrac{2}{3}(0.426)(11.3\times10^{-2})^2(67.86)^2\\KE_{rotational}=8.35\rm\; J[/tex]
Hence, the rate at which the soccer ball was rotating at the base of the hill is 67.86 rad/s and the rotational kinetic energy is 8.35 J.
Learn more about the kinetic energy here;
https://brainly.com/question/25959744
