10.4 Moment of inertia and rotational kinetic energy  (Page 3/7)

Moment of inertia of a system of particles

Six small washers are spaced 10 cm apart on a rod of negligible mass and 0.5 m in length. The mass of each washer is 20 g. The rod rotates about an axis located at 25 cm, as shown in [link] . (a) What is the moment of inertia of the system? (b) If the two washers closest to the axis are removed, what is the moment of inertia of the remaining four washers? (c) If the system with six washers rotates at 5 rev/s, what is its rotational kinetic energy?

Strategy

1. We use the definition for moment of inertia for a system of particles and perform the summation to evaluate this quantity. The masses are all the same so we can pull that quantity in front of the summation symbol.
2. We do a similar calculation.
3. We insert the result from (a) into the expression for rotational kinetic energy.

Solution

1. $I={\displaystyle \sum _j{m}_j{r}_j^2}=(0.02\text{kg})(2\times {(0.25\text{m})}^2+2\times {(0.15\text{m})}^2+2\times {(0.05\text{m})}^2)=0.0035\text{kg}·{\text{m}}^2$ .
2. $I={\displaystyle \sum _j{m}_j{r}_j^2}=(0.02\text{kg})(2\times {(0.25\text{m})}^2+2\times {(0.15\text{m})}^2)=0.0034\text{kg}·{\text{m}}^2$ .
3. $K=\frac{1}{2}I{\omega }^2=\frac{1}{2}(0.0035\text{kg}·{\text{m}}^2)(5.0\times 2\pi \text{rad}\text{/}{\text{s})}^2=1.73\text{J}$ .

Significance

We can see the individual contributions to the moment of inertia. The masses close to the axis of rotation have a very small contribution. When we removed them, it had a very small effect on the moment of inertia.

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Calculating helicopter energies

A typical small rescue helicopter has four blades: Each is 4.00 m long and has a mass of 50.0 kg ( [link] ). The blades can be approximated as thin rods that rotate about one end of an axis perpendicular to their length. The helicopter has a total loaded mass of 1000 kg. (a) Calculate the rotational kinetic energy in the blades when they rotate at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter when it flies at 20.0 m/s, and compare it with the rotational energy in the blades.

Strategy

Rotational and translational kinetic energies can be calculated from their definitions. The wording of the problem gives all the necessary constants to evaluate the expressions for the rotational and translational kinetic energies.

Solution

1. The rotational kinetic energy is
   
   $K=\frac{1}{2}I{\omega }^2.$
   
   We must convert the angular velocity to radians per second and calculate the moment of inertia before we can find K . The angular velocity $\omega$ is
   
   $\omega =\frac{300\text{rev}}{1.00\text{min}}\frac{2\pi \text{rad}}{\text{1 rev}}\frac{1.00\text{min}}{60.0\text{s}}=31.4\frac{\text{rad}}{\text{s}}.$
   
   The moment of inertia of one blade is that of a thin rod rotated about its end, listed in [link] . The total I is four times this moment of inertia because there are four blades. Thus,
   
   $I=4\frac{M{l}^2}{3}=4\times \frac{(50.0\text{kg}){(4.00\text{m})}^2}{3}=1067.0\text{kg}·{\text{m}}^2.$
   
   Entering $\omega$ and I into the expression for rotational kinetic energy gives
   
   $K=0.5(1067\text{kg}·{\text{m}}^2){\text{(31.4 rad/s)}}^2=5.26\times {10}^5\text{J}\text{.}$
2. Entering the given values into the equation for translational kinetic energy, we obtain
   
   $K=\frac{1}{2}m{v}^2=(0.5)(1000.0\text{kg}){(20.0\text{m/s})}^2=2.00\times {10}^5\text{J}.$
   
   To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is
   
   $\frac{2.00\times {10}^5\text{J}}{5.26\times {10}^5\text{J}}=0.380.$

Significance

The ratio of translational energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter is in its spinning blades.

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