10.4 Moment of inertia and rotational kinetic energy

Energy in a boomerang

A person hurls a boomerang into the air with a velocity of 30.0 m/s at an angle of $40.0\text{°}$ with respect to the horizontal ( [link] ). It has a mass of 1.0 kg and is rotating at 10.0 rev/s. The moment of inertia of the boomerang is given as $I=\frac{1}{12}m{L}^2$ where $L=0.7\text{m}$ . (a) What is the total energy of the boomerang when it leaves the hand? (b) How high does the boomerang go from the elevation of the hand, neglecting air resistance?

Strategy

We use the definitions of rotational and linear kinetic energy to find the total energy of the system. The problem states to neglect air resistance, so we don’t have to worry about energy loss. In part (b), we use conservation of mechanical energy to find the maximum height of the boomerang.

Solution

1. Moment of inertia: $I=\frac{1}{12}m{L}^2=\frac{1}{12}(1.0\text{kg})(0.7{\text{m})}^2=0.041\text{kg}·{\text{m}}^2$ .
   Angular velocity: $\omega =(10.0\text{rev}\text{/}\text{s})(2\pi )=62.83\text{rad}\text{/}\text{s}$ .
   The rotational kinetic energy is therefore
   
   $K_{\text{R}}=\frac{1}{2}(0.041\text{kg}·{\text{m}}^2){(62.83\text{rad}\text{/}\text{s})}^2=80.93\text{J}.$
   
   The translational kinetic energy is
   
   $K_{\text{T}}=\frac{1}{2}m{v}^2=\frac{1}{2}(1.0\text{kg})(30.0\text{m}\text{/}\text{s}{)}^2=450.0\text{J}.$
   
   Thus, the total energy in the boomerang is
   
   $K_{\text{Total}}=K_{\text{R}}+K_{\text{T}}=80.93+450.0=530.93\text{J}.$
2. We use conservation of mechanical energy. Since the boomerang is launched at an angle, we need to write the total energies of the system in terms of its linear kinetic energies using the velocity in the x - and y -directions. The total energy when the boomerang leaves the hand is
   
   $E_{\text{Before}}=\frac{1}{2}m{v}_x^2+\frac{1}{2}m{v}_y^2+\frac{1}{2}I{\omega }^2.$
   
   The total energy at maximum height is
   
   $E_{\text{Final}}=\frac{1}{2}m{v}_x^2+\frac{1}{2}I{\omega }^2+mgh.$
   
   By conservation of mechanical energy, $E_{\text{Before}}=E_{\text{Final}}$ so we have, after canceling like terms,
   
   $\frac{1}{2}m{v}_y^2=mgh.$
   
   Since $v_y=30.0\text{m}\text{/}\text{s}(\text{sin}40\text{°})=19.28\text{m}\text{/}\text{s}$ , we find
   
   $h=\frac{{(19.28\text{m}\text{/}\text{s})}^2}{2(9.8\text{m}\text{/}{\text{s}}^2)}=18.97\text{m}.$

Significance

In part (b), the solution demonstrates how energy conservation is an alternative method to solve a problem that normally would be solved using kinematics. In the absence of air resistance, the rotational kinetic energy was not a factor in the solution for the maximum height.

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