Respuesta :
Answer:
L = 30 m
Explanation:
As per mechanical energy conservation law we can say that total kinetic energy of the hoop is equal to the total gravitational potential energy at the top
So here we can say for initial total kinetic energy as
[tex]KE = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2[/tex]
for pure rolling we will have
[tex]v = r\omega[/tex]
also for large hoop we will have
[tex]I = mR^2[/tex]
now we have
[tex]KE = \frac{1}{2}mR^2\omega^2 + \frac{1}{2}(mR^2)\omega^2[/tex]
[tex]KE = mR^2\omega^2[/tex]
[tex]KE = (3.8)(1.3)^2(6.7)^2[/tex]
[tex]KE = 288.3 J[/tex]
now for gravitational potential energy we can say
[tex]U = mg(Lsin\theta)[/tex]
now by energy conservation we have
[tex]288.3 = (3.8)(9.81)(L sin15)[/tex]
[tex]L = 30 m[/tex]
The horizontal distance traveled by the hoop is approximately 30 m.
Total kinetic energy of the ball
The total kinetic energy of the ball is the sum of the translational and rotational kinetic of the ball.
K.E(total) = K.E(trans) + K.E(rotational)
[tex]K.E(t) = \frac{1}{2}mv^2 + \frac{1}{2} I \omega ^2\\\\K.E(t) = \frac{1}{2}mv^2 + \frac{1}{2} (mR^2) \omega ^2\\\\K.E(t) = \frac{1}{2}m(\omega R)^2 + \frac{1}{2} (mR^2) \omega ^2\\\\K.E(t) = m\omega ^2R^2\\\\K.E(t) = 3.8 \times (6.7)^2 \times (1.3)^2\\\\K.E(t) = 288.28 \ J[/tex]
Conservation of energy
The horizontal distance traveled by the hoop is calculated as follows;
K.E = P.E
288.28 = mg(Lsinθ)
288.28 = (3.8 x 9.8) x (L) x sin(15)
288.28 = 9.63 L
L = 30 m
Thus, the horizontal distance traveled by the hoop is approximately 30 m.
Learn more about conservation of energy here: https://brainly.com/question/166559
