10.4 Moments of inertia and rotational kinetic energy - University Physics Part 1 | Open Stax (2023)

$$K=\frac{1}{2}（{\displaystyle \underset{J}{S}{\mathrm{Medium}}_J{\mathrm{right}}_J^2}）{\mathrm{oh}}^2。$$

10.16

$$\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_J{\mathrm{right}}_J^2}。$$

10.17

$$K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^2。$$

10.18

The moment of inertia of the particle system

On a rod of negligible mass and length 0.5 m, 6 small rings each having a mass of 20 g are placed at a distance of 10 cm, and the rod is rotated about an axis located at 25 cm as shownFigure 10.19. (a) What is the moment of inertia of the system? (b) If the two rings closest to the axis are removed, what are the moments of inertia of the remaining four rings? (c) If the six-ring system is rotating at 5 revolutions per minute, what is its rotational kinetic energy?

Lectra10.19 A rod of negligible mass has six rings spaced 10 cm apart and rotates about a vertical axis.

strategy

1. We estimate this quantity using the definition of the moment of inertia of the particle system and summing. All masses are the same, so we can put that quantity in front of the addition sign.
2. We also do similar calculations.
3. We substitute the result of (a) into the expression for rotational kinetic energy.

solution

1. $\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_J{\mathrm{right}}_J^2}=（\mathrm{0,02}\text{Kilogram}）（2\times {（\mathrm{0,25}\text{Medium}）}^2+2\times {（\mathrm{0,15}\text{Medium}）}^2+2\times {（\mathrm{0,05}\text{Medium}）}^2）=\mathrm{0,0035}\text{Kilogram}·{\text{Medium}}^2$。
2. $\mathrm{and}={\displaystyle \underset{J}{S}{\mathrm{Medium}}_J{\mathrm{right}}_J^2}=（\mathrm{0,02}\text{Kilogram}）（2\times {（\mathrm{0,25}\text{Medium}）}^2+2\times {（\mathrm{0,15}\text{Medium}）}^2）=\mathrm{0,0034}\text{Kilogram}·{\text{Medium}}^2$。
3. $K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^2=\frac{1}{2}（\mathrm{0,0035}\text{Kilogram}·{\text{Medium}}^2）（\mathrm{5,0}\times 2\mathrm{PI}\text{rad}\text{/}{\text{S}）}^2=\mathrm{1,73}\text{J}$。

significance

We can see the individual contribution to the moment of inertia. The contribution of masses close to the axis of rotation is very small. When we remove them, it has very little effect on the moment of inertia.

rotational energy

1. Determine if energy or work is involved in the rotation.
2. Identify systems of interest. Often a sketch will help.
3. Analyze the situation to determine what types of work and energy are involved.
4. Mechanical energy is conserved if no energy is lost due to friction and other non-conservative forces, ie:$K_{\text{and}}+U_{\text{and}}=K_{\text{F}}+U_{\text{F}}$。
5. When non-conservative forces are present, mechanical energy is not conserved and other forms of energy (such as heat and light) can enter or leave the system. Determine what they are and calculate them if necessary.
6. Eliminate terminology wherever possible to simplify algebra.
7. Evaluate the numerical solution to see if it makes sense for the physical situation posed in the problem statement.

Calculating the Energy of a Helicopter

typical little rescuehelicopterThere are four blades: each blade is 4.00 m long and has a mass of 50.0 kg (Figure 10.21). A blade can be approximated as a thin rod that rotates at one end about an axis perpendicular to its length. The helicopter has a total load mass of 1000 kg. (a) Calculate the kinetic energy of rotation when the blade rotates at 300 rpm. (b) Calculate the translational kinetic energy of the helicopter flying at 20.0 m/s and compare it to the rotational energy of the blades.

Lectra10.21 (a) Sketch of a quadrocopter. (b) Helicopter water rescue operations operated by Westpac Rescue Helicopter Services Auckland. (Image credit b: "111 Emergency"/Flickr job changes)

strategy

Rotational and translational energies can be calculated according to their definitions. The formulation of the problem provides all the necessary constants for evaluating the rotational and translational kinetic energy expressions.

solution

1. The rotational kinetic energy is

   $$K=\frac{1}{2}\mathrm{and}{\mathrm{oh}}^2。$$

   We need to convert the angular velocity to radians per second and calculate the moment of inertia to find itK. angular velocity$\mathrm{oh}$yes

   $$\mathrm{oh}=\frac{300\text{Rotating speed}}{\mathrm{1,00}\text{minute}}\frac{2\mathrm{PI}\text{rad}}{\text{1 district}}\frac{\mathrm{1,00}\text{minute}}{\mathrm{60,0}\text{S}}=31.4\frac{\text{rad}}{\text{S}}。$$

   The moment of inertia of a blade is equal to the moment of rotation of a thin rod at its end as followsFigure 10.20. totalandis four times the moment of inertia because there are four blades. so,

   $$\mathrm{and}=4\frac{\mathrm{Medium}{L}^2}{3}=4\times \frac{（\mathrm{50,0}\text{Kilogram}）{（\mathrm{4:00\; AM}\text{Medium}）}^2}{3}=\mathrm{1067,0}\text{Kilogram}·{\text{Medium}}^2。$$

   Nurse$\mathrm{oh}$UandIn the expression for rotational kinetic energy

   $$K=\mathrm{0,5}（1067\text{Kilogram}·{\text{Medium}}^2）{\text{(31.4 rad/s)}}^2=5.26\times {10}^5\text{J}\text{。}$$

2. Substituting the given values ​​into the translational kinetic energy equation, we get

   $$K=\frac{1}{2}\mathrm{Medium}{v}^2=（\mathrm{0,5}）（\mathrm{1000,0}\text{Kilogram}）{（20.0\text{Mrs}）}^2=\mathrm{2,00}\times {10}^5\text{J}。$$

   To compare kinetic energies, we take the ratio of translational kinetic energy to rotational kinetic energy. This ratio is

   $$\frac{\mathrm{2,00}\times {10}^5\text{J}}{5.26\times {10}^5\text{J}}=\mathrm{0,380}。$$

significance

The ratio of translational kinetic energy to rotational kinetic energy is only 0.380. This ratio tells us that most of the kinetic energy of the helicopter comes from the rotating blades.

boomerang energy

roll oneboomerangIn air at a velocity of 30.0 m/s at an angle of$\mathrm{40,0}\text{°}$relative to the horizontal plane (Figure 10.22). It has a mass of 1.0 kg and spins at 10.0 revolutions per minute. The moment of inertia of the boomerang is$\mathrm{and}=\frac{1}{12}\mathrm{Medium}{L}^2$Where$L=\mathrm{0,7}\text{Medium}$. (a) What is the total energy of the boomerang when it leaves the hand? (b) Neglecting air resistance, how high does the boomerang rise from the height of the fist?

Lectra10.22 The boomerang is thrown into the air at an initial angle$40\text{°}$。

strategy

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

solution

1. Moment of inertia:$\mathrm{and}=\frac{1}{12}\mathrm{Medium}{L}^2=\frac{1}{12}（\mathrm{1,0}\text{Kilogram}）（\mathrm{0,7}{\text{Medium}）}^2=\mathrm{0,041}\text{Kilogram}·{\text{Medium}}^2$。
   Angular velocity:$\mathrm{oh}=（10.0\text{Rotating speed}\text{/}\text{S}）（2\mathrm{PI}）=\mathrm{62,83}\text{rad}\text{/}\text{S}$。
   Therefore the rotational kinetic energy is

   $$K_{\text{right}}=\frac{1}{2}（\mathrm{0,041}\text{Kilogram}·{\text{Medium}}^2）{（\mathrm{62,83}\text{rad}\text{/}\text{S}）}^2=\mathrm{80,93}\text{J}。$$

   Translational kinetic energy is

   $$K_{\text{time}}=\frac{1}{2}\mathrm{Medium}{v}^2=\frac{1}{2}（\mathrm{1,0}\text{Kilogram}）（30.0\text{Medium}\text{/}\text{S}{）}^2=\mathrm{450,0}\text{J}。$$

   Therefore, the total energy of the boomerang is:

   $$K_{\text{total}}=K_{\text{right}}+K_{\text{time}}=\mathrm{80,93}+\mathrm{450,0}=\mathrm{530,93}\text{J}。$$

2. We use the law of conservation of mechanical energy. Since the boomerang is launched at an angle, we must write the total energy of the system in terms of its linear kinetic energy, using the velocityX- inj- Travel instructions. The total energy of the boomerang when it leaves the hand is

   $${\mathrm{Second}}_{\text{for}}=\frac{1}{2}\mathrm{Medium}{v}_X^2+\frac{1}{2}\mathrm{Medium}{v}_j^2+\frac{1}{2}\mathrm{and}{\mathrm{oh}}^2。$$

   The total energy at the highest altitude is

   $${\mathrm{Second}}_{\text{Last}}=\frac{1}{2}\mathrm{Medium}{v}_X^2+\frac{1}{2}\mathrm{and}{\mathrm{oh}}^2+\mathrm{Medium}GH。$$

   By saving mechanical energy,${\mathrm{Second}}_{\text{for}}={\mathrm{Second}}_{\text{Last}}$So, after removing similar conditions, we

   $$\frac{1}{2}\mathrm{Medium}{v}_j^2=\mathrm{Medium}GH。$$

   from$v_j=30.0\text{Medium}\text{/}\text{S}（\text{Slam}40\text{°}）=19.28\text{Medium}\text{/}\text{S}$, we discover

   $$H=\frac{{（19.28\text{Medium}\text{/}\text{S}）}^2}{2（9.8\text{Medium}\text{/}{\text{S}}^2）}=18.97\text{Medium}。$$

significance

In part (b), the solution shows how energy conservation can be an alternative to solving problems that are usually solved using kinematics. In the absence of air resistance, rotational kinetic energy plays no role in the maximum altitude solution.