Answer to Question #216749 in Mechanics | Relativity for Jessy

Question

1. Compute the rotational Kinetic energy of a 25 kg wheel rotating at 6 rev/s if the radius of
gyration of the wheel is 22 cm.
Note: the radius of gyration is the point where the axis of rotation passes through. This axis is a
result of the parallel axis theorem. In this problem, since it is given, you can already compute for
what is being asked for.
2. To crack a certain nut in a nutcracker, forces with magnitudes of at least 40 N must act on its
shell from both sides. For the nutcracker of Fig. 12-34, with distances

Answer 1

The general expression for Rotational Kinetic Energy of a body is:

K.E = $\frac{1}{2}I\omega^2$

• "I" refers to Moment Of Inertia ,  represents the angular velocity.
• Moment Of Inertia can also be represented as product of mass and square of radius of gyration.

K.E = $\frac{1}{2}(mk^2)\omega^2$

Putting available values in SI units:

K.E = $\frac{1}{2}[25 \times (\frac{22}{100})^2](6 \times 2\pi)^2$

K.E = $\frac{1}{2}[25 \times (\frac{484}{10000})](6 \times 2\pi)^2$

K.E = $\frac{1}{2}[1.21](6 \times 2\pi)^2$

K.E = $\frac{1}{2}[1.21](36 \times 4\pi^2)$

K.E = 859.83 joule

K.E ~ 860 joule.

So, Rotational Kinetic Energy is approximately 860 Joules.