Answer:
The moment of inertia is 1.01 kg m²
Explanation:
Given:
θ = 8.75 revolutions = 2π * 8.75 = 54.98 rad
t = time = 13 s
Ek = kinetic energy = 36 J
The angular acceleration is equal to:
[tex]\alpha =\frac{\theta }{0.5t^{2} } =\frac{54.98}{0.5*13^{2} } =0.65rad/s^{2}[/tex]
The angular velocity is:
w = αt = 0.65 * 13 = 8.45 rad/s
The moment of inertia is:
[tex]I=\frac{2E_{k} }{w^{2} } =\frac{2*36}{8.45^{2} } =1.01kgm^{2}[/tex]
The moment of inertia of the wheel for an axis through its center is equal to 1.01 [tex]kgm^2[/tex].
Given the following data:
Conversion:
Final angular displacement = 8.75 revs to rad = [tex]8.75 \times 2\pi = 54.98\;rad[/tex]
To calculate the moment of inertia of the wheel for an axis through its center:
Mathematically, the final angular speed is given by this formula:
[tex]\theta_f - \theta_i = \omega_i t + \frac{1}{2} \alpha t\\\\\theta_f - \theta_i = \omega_i t + \frac{1}{2} (\omega_f -\omega_i) t\\\\\theta_f - 0 = 0t + \frac{1}{2} (\omega_f -0) t\\\\\theta_f = \frac{1}{2} \omega_f t\\\\\omega_f = \frac{ 2 \theta_f}{t}[/tex]
Substituting the given parameters into the formula, we have;
[tex]\omega_f = \frac{ 2 \times 54.98}{13.0} \\\\\omega_f = \frac{ 109.96}{13.0} \\\\\omega_f =8.46\;rad/s[/tex]
Now, we can calculate the moment of inertia of the wheel by using this formula:
[tex]K.E = \frac{1}{2} I\omega_f^2\\\\I=\frac{2K.E}{\omega_f^2} \\\\I=\frac{2 \times 36.0}{8.46^2} \\\\I=\frac{72.0}{71.57}[/tex]
Moment of inertia, I = 1.01 [tex]kgm^2[/tex]
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