(A) [tex]4.2\cdot 10^5 J[/tex]
The energy stored by the system is given by
[tex]E=Pt[/tex]
where
P is the power provided
t is the time elapsed
In this case, we have
P = 60 kW = 60,000 W is the power
t = 7 is the time
Therefore, the energy stored by the system is
[tex]E=(60,000 W)(7 s)=4.2\cdot 10^5 J[/tex]
(B) 4830 rad/s
The rotational energy of the wheel is given by
[tex]E=\frac{1}{2}I \omega^2[/tex] (1)
where
[tex]I[/tex] is the moment of inertia
[tex]\omega[/tex] is the angular velocity
The moment of inertia of the wheel is
[tex]I=\frac{1}{2}MR^2=\frac{1}{2}(5 kg)(0.12 m)^2=0.036 kg m^2[/tex]
where M is the mass and R the radius of the wheel.
We also know that the energy provided is
[tex]E=4.2\cdot 10^5 J[/tex]
So we can rearrange eq.(1) to find the angular velocity:
[tex]\omega=\sqrt{\frac{2E}{I}}=\sqrt{\frac{2(4.2\cdot 10^5 J)}{0.036 kg m^2}}=4830 rad/s[/tex]
(C) [tex]2.8\cdot 10^6 m/s^2[/tex]
The centripetal acceleration of a point on the edge is given by
[tex]a=\omega^2 R[/tex]
where
[tex]\omega=4830 rad/s[/tex] is the angular velocity
R = 0.12 m is the radius of the wheel
Substituting, we find
[tex]a=(4830 rad/s)^2 (0.12 m)=2.8\cdot 10^6 m/s^2[/tex]
