Answer:
h=12.41m
Explanation:
N=392
r=0.6m
w=24 rad/s
[tex]I=0.8*m*r^{2}[/tex]
So the weight of the wheel is the force N divide on the gravity and also can find momentum of inertia to determine the kinetic energy at motion
[tex]N=m*g\\m=\frac{N}{g}\\m=\frac{392N}{9.8\frac{m}{s^{2}}}[/tex]
[tex]m=40kg[/tex]
moment of inertia
[tex]M_{I}=0.8*40.0kg*(0.6m} )^{2}\\M_{I}=11.5 kg*m^{2}[/tex]
Kinetic energy of the rotation motion
[tex]K_{r}=\frac{1}{2}*I*W^{2}\\K_{r}=\frac{1}{2}*11.52kg*m^{2}*(24\frac{rad}{s})^{2}\\K_{r}=3317.76J[/tex]
Kinetic energy translational
[tex]K_{t}=\frac{1}{2}*m*v^{2}\\v=w*r\\v=24rad/s*0.6m=14.4 \frac{m}{s}\\K_{t}=\frac{1}{2}*40kg*(14.4\frac{m}{s})^{2}\\K_{t}=4147.2J[/tex]
Total kinetic energy
[tex]K=3317.79J+4147.2J\\K=7464.99J[/tex]
Now the work done by the friction is acting at the motion so the kinetic energy and the work of motion give the potential work so there we can find height
[tex]K-W=E_{p}\\7464.99-2600J=m*g*h\\4864.99J=m*g*h\\h=\frac{4864.99J}{m*g}\\h=\frac{4864.99J}{392N}\\h=12.41m[/tex]
The height above the bottom of the hill rolled up by the wheel is 15.1 m.
The given parameters;
The mass of wheel, M, is calculated as follows;
[tex]W = Mg\\\\M = \frac{W}{g} = \frac{392}{9.8} = 40 \ kg[/tex]
The work-done by friction on the wheel is change in the mechanical energy of the wheel;
[tex]mgh - K.E_r = W_f\\\\mgh = W_f + K.E_r\\\\mgh = W_f + \frac{1}{2} I\omega^2 \\\\mgh = W_f + \frac{1}{2} (0.8MR^2 \omega ^2)\\\\h = \frac{W_f \ + \ 0.4MR^2 \omega ^2}{mg} \\\\h = \frac{2600 \ \ + \ \ 0.4 \times (40 )\times 0.6^2 \times (24)^2}{392} \\\\h = 15.1 \ m[/tex]
Thus, the height above the bottom of the hill rolled up by the wheel is 15.1 m.
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