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Rotational inertia and moment of inertia

Before we can consider the rotation of anything other than a point mass like the one in [link] , we must extend the idea of rotational inertia to all types of objects. To expand our concept of rotational inertia, we define the moment of inertia     I size 12{I} {} of an object to be the sum of mr 2 size 12{ ital "mr" rSup { size 8{2} } } {} for all the point masses of which it is composed. That is, I = mr 2 size 12{I= Sum {} ital "mr" rSup { size 8{2} } } {} . Here I size 12{I} {} is analogous to m size 12{m} {} in translational motion. Because of the distance r size 12{r} {} , the moment of inertia for any object depends on the chosen axis. Actually, calculating I size 12{I} {} is beyond the scope of this text except for one simple case—that of a hoop, which has all its mass at the same distance from its axis. A hoop’s moment of inertia around its axis is therefore MR 2 size 12{ ital "MR" rSup { size 8{2} } } {} , where M size 12{M} {} is its total mass and R size 12{R} {} its radius. (We use M size 12{M} {} and R size 12{R} {} for an entire object to distinguish them from m size 12{m} {} and r size 12{r} {} for point masses.) In all other cases, we must consult [link] (note that the table is piece of artwork that has shapes as well as formulae) for formulas for I size 12{I} {} that have been derived from integration over the continuous body. Note that I size 12{I} {} has units of mass multiplied by distance squared ( kg m 2 size 12{"kg" cdot "m" rSup { size 8{2} } } {} ), as we might expect from its definition.

The general relationship among torque, moment of inertia, and angular acceleration is

net τ = size 12{τ=Iα} {}

or

α = net τ I , size 12{α= { { ital "net"τ} over {I} } ","} {}

where net τ size 12{τ} {} is the total torque from all forces relative to a chosen axis. For simplicity, we will only consider torques exerted by forces in the plane of the rotation. Such torques are either positive or negative and add like ordinary numbers. The relationship in τ = α = net τ I size 12{τ=Iα,`````α= { { ital "net"τ} over {I} } } {} is the rotational analog to Newton’s second law and is very generally applicable. This equation is actually valid for any torque, applied to any object, relative to any axis.

As we might expect, the larger the torque is, the larger the angular acceleration is. For example, the harder a child pushes on a merry-go-round, the faster it accelerates. Furthermore, the more massive a merry-go-round, the slower it accelerates for the same torque. The basic relationship between moment of inertia and angular acceleration is that the larger the moment of inertia, the smaller is the angular acceleration. But there is an additional twist. The moment of inertia depends not only on the mass of an object, but also on its distribution of mass relative to the axis around which it rotates. For example, it will be much easier to accelerate a merry-go-round full of children if they stand close to its axis than if they all stand at the outer edge. The mass is the same in both cases; but the moment of inertia is much larger when the children are at the edge.

Problem-solving strategy for rotational dynamics

  1. Examine the situation to determine that torque and mass are involved in the rotation . Draw a careful sketch of the situation.
  2. Determine the system of interest .
  3. Draw a free body diagram . That is, draw and label all external forces acting on the system of interest.
  4. Apply net τ = α = net τ I size 12{τ=Iα,```α= { { ital "net"τ} over {I} } } {} , the rotational equivalent of Newton’s second law, to solve the problem . Care must be taken to use the correct moment of inertia and to consider the torque about the point of rotation.
  5. As always, check the solution to see if it is reasonable .

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
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A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
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Can you compute that for me. Ty
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what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
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A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
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you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
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Magreth
progressive wave
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Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
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Source:  OpenStax, Unit 8 - rotational motion. OpenStax CNX. Feb 22, 2016 Download for free at https://legacy.cnx.org/content/col11970/1.1
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