Answer:
a) [tex]\omega_{f} = \sqrt{\omega_{o}^{2}+ \frac{2\cdot F \cdot L}{k\cdot m} }[/tex], b) [tex]\dot W = F\cdot r \cdot \omega_{o}[/tex]
Explanation:
a) The final rotational speed is:
[tex]\omega_{f} = \sqrt{\omega_{o}^{2}+ 2\cdot r \cdot \alpha\cdot L}[/tex]
But:
[tex]T = I\cdot \alpha[/tex]
[tex]\alpha = \frac{T}{I}[/tex]
[tex]\alpha = \frac{F\cdot r}{k\cdot m \cdot r^{2}}[/tex]
[tex]\alpha = \frac{F}{k\cdot m \cdot r}[/tex]
Lastly, the angular acceleration is replace in the expression for the final angular velocity:
[tex]\omega_{f} = \sqrt{\omega_{o}^{2}+ \frac{2\cdot F \cdot L}{k\cdot m} }[/tex]
b) The power is rate of change of work in terms of time:
[tex]W = F\cdot r\cdot \theta[/tex]
Instantaneous power is obtained by differentiating the previous expression with respect to time:
[tex]\dot W = F\cdot r \cdot \omega(t)[/tex]
The instantaneous power delivered to the wheel at t = 0 is:
[tex]\dot W = F\cdot r \cdot \omega_{o}[/tex]
A) The final rotational speed ω_final of the wheel is; ω_f = √((ω_i)² + 2(FL/(kmr²)
B) The instantaneous power P delivered to the wheel via the force F at time t=0 is; P = Frω_i
From rotational kinematics, the formula for the final angular velocity is;
ω_f = √((ω_i)² + 2αθ)
where;
α is angular acceleration
θ = L/r. Thus;
ω_f = √((ω_i)² + 2α(L/r))
Now, α = T/I
Where;
I is moment of inertia = kmr²
T is torque = F*r
Thus;
α = (F * r)/(kmr²)
α = F/(kmr)
ω_f = √((ω_i)² + 2(F/(kmr))(L/r))
ω_f = √((ω_i)² + 2(FL/(kmr²)
B) Formula for instantaneous power is;
P = Fv
where at t = 0; v = rω_i
Thus;
P = Frω_i
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