Calculate the rotational energy of a segment given the mass of the segment is 2.2 kg the moment of inertia is 0.57 kg-m², and the angular velocity is 25 rad/s. 178 J 356 J 102 J 7.13 J
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Step 1: The formula for rotational kinetic energy is given by: $KE_{rot} = \frac{1}{2}I\omega^2$ where: $KE_{rot}$ is the rotational kinetic energy $I$ is the moment of inertia $\omega$ is the angular velocity Show more…
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Calculate the rotational energy of a segment, given mass of the segment is 5.68 kg, moment of inertia is 0.93 kg-m2, and angular velocity is 29 rad/s.
Sri K.
Given, $M=20 \mathrm{~kg}$ $$ \begin{gathered} \omega=100 \mathrm{rad} / \mathrm{s} \\ R=0.25 \mathrm{~m} \end{gathered} $$ Moment of inertia of the solid cylinder about its axis of symmetry. $$ \begin{gathered} \quad I=\frac{1}{2} M R^{2} \\ =\frac{1}{2} \times 20 \times(0.25)^{2} \\ =10 \times 0.0625 \\ =0.625 \mathrm{~kg}-\mathrm{m}^{2} \end{gathered} $$ Kinetic energy associated with the rotation of the cylinder is given by $$ K=\frac{1}{2} / \omega^{2} $$ $$ \begin{aligned} &=\frac{1}{2} \times 0.625 \times(100)^{2} \\ &=0.3125 \times 10000 \\ &=3125 \mathrm{~J} \end{aligned} $$ Angular momentum, $L=/ \omega$ $$ \begin{aligned} &=0.625 \times 100 \\ &=62.5 \mathrm{~J}-\mathrm{s} \end{aligned} $$
Gravitation
Round 2
What is the rotational kinetic energy of an object that has a moment of inertia of $0.280 \mathrm{~kg} \cdot \mathrm{m} 2^{0.280 \mathrm{~kg} \cdot \mathrm{m}^{2}}$ about the axis of rotation when its angular speed is $4.00 \mathrm{rad} / \mathrm{s}$ ? Example $8-2$
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