Answer:
The value of h is 14 meters
Explanation:
Given that,
Weight of the wheel, F = 392 N
Angular velocity of the wheel, [tex]\omega=25\ rad/s[/tex]
Radius of the wheel, r = 0.6 m
The moment of inertia about its rotation axis is, [tex]I=0.8\ mr^2[/tex]
Work done by the wheel, W = 2600 J
Initially, the wheel has both translational and rotational kinetic energy such that,
[tex]E_r+E_t-W=PE\\\\\dfrac{1}{2}I\omega^2+\dfrac{1}{2}mv^2-W=mgh[/tex]
Use [tex]v=r\omega[/tex]
[tex]\dfrac{1}{2}\times 0.8mr^2\omega^2+\dfrac{1}{2}m\omega^2r^2-W=mgh\\\\\dfrac{1}{2}\times 0.8\times \dfrac{392}{9.8}\times 0.6^2\times (25)^2+\dfrac{1}{2}\times \dfrac{392}{9.8}\times (25)^2\times (0.6)^2-2600=mgh\\\\h=\dfrac{5500}{mg}\\\\h=\dfrac{5500}{40\cdot9.8}\\\\h=14.03\ m[/tex]
So, the value of h is 14 meters.
