1. Rotational Inertia Of Bearing Diameter
2. Moment Of Inertia Of Bearing

Using Rotation

The experimental data demonstrated that the rotational inertia of three rotors in the model A is 6.293×10 -5 kgm 2, 1.074×10 -4 kgm 2 and 2.081×10 -4 kgm 2 respectively, and the corresponding. The rotational inertia of the bar is given by: I B = (½) m o (l 2 + w 2) where m o is the mass of the bar, l its length, and w its width. The rotational inertia of the cylindrical masses is given by; I C = 2(½) Mp^2 + 2MR^2 where M is the mass of one cylinder, and? Is its diameter. Inertia, J Bearings, T 0, b MASSACHUSETTS INSTITUTE OF TECHNOLOGY Department of Mechanical Engineering 2.004 Dynamics and Control II Fall 2007 Lecture 2 Solving the Equation of Motion Goals for today. Modeling of the 2.004 Lab’s rotational system. Analytical solution of the equation of motion for a 1st–order system using the time domain.

... or putting a new spin on things. You have heard the phrase 'spinning out of control'. In fact, you can GAIN control by the action of a spinning object.

Key Terms

centrifugal force
centrifugal governor
centrifuge
centripetal acceleration
centripetal force
feedback (negative & positive)
geosynchronous orbit
inertial switch
moment of inertia

Centrifugal Governors ... the classic controlling device

Rotational Inertia Of Bearing Diameter

The governor is a component in an engine which maintains a certain RPM based on the setting of your throttle. If your throttle is set for a certain speed (say 1000 RPM) and you place a load on the engine which begins to slow it down, the governor will compensate by opening the carburetor and allowing extra gas and air in. Most tractors and small engines (like on your lawnmower) have a governor between the throttle and the carburetor.

Below is a picture of an early governor developed by James Watt in his steam engine

Watt Centrifugal Governor Courtesy Wikimedia Commons

As the flywheel slows in rotational speed, the arms lower. This opens the throttle valve to release more steam … making the device spin faster. Likewise, if the governor spins too fast, the arms rise up to close the throttle valve. This cuts the steam supply and the device spins at a slower rate. This makes the governor a perfect feedback mechanism to keep the shaft spinning at a nearly constant rate.

A similar device is found in elevators. If there was a problem and the car suddenly accelerates beyond safety limits ... a centrifugal governor activates a safety brake. Some rental trucks and school buses have these devices built in to avoid excess speeds.

Some small engines (50 cc and under) have a device that limits the RPMs in a similar manner. They have a small device that is linked to the throttle that is activated by air blown off the flywheel. If the RPMs rise too high, the flywheel spins faster and makes more draft ... which in turn pushes on a small 'sail valve' which turns down the mixture to the carburetor. Modern engines may control the RPMs by 'electronically' counting the speed and sending the information to a microprocessor.

All cars need transmissions. This is because the engine spins at a small range of RPM but the wheels need a much larger range of speeds (and torque). There is a governor in the automatic transmission of your car which tells the system when to change gear ratios. As the rotation speed increases, the spring retracted governor will either allow more or less hydraulic fluid through the system to open/close bands and clutches ... allowing gears to spin at different rates and directions.

Starting your lawn mower

Have you ever wondered what happens when you pull the rope to start your lawn mower? Just like the very first cars, you have to get the motor turning by hand to get it started. Although we will cover the way an engine works later in this class ... for now, let's just see how the manual starter works.

As you know, you have to pull on a rope and the engine turns as a result. But how does the starter engage the engine? The starter assembly attaches to the top of the engine's flywheel. When you pull on the rope, small 'swing levers' jet out and engage cogs on the inside hub of the flywheel. Normally, these swing levers are held inward by small springs. Once you are done pulling on the rope (and the engine starts up), the swing levers retract to their rest position.

Similarly, my chain saw uses a centrifugal clutch to transfer the energy of the engine's spinning crankshaft to a drum (that moves the chain). As the engine revs up, friction plates move outward and make contact with the drum. At low engine speeds, retraction springs release this contact and the chain stops moving.

This same idea is used in a centrifugal switch. Some induction motors use an extra set of windings to kick-start the motor when it initially starts up (to overcome inertia). Once the motor reaches normal operational speed, power to these extra windings is no longer needed, so you need a way of switching it off. Can you see how the ideas used here can be used to break an electrical connection when the shaft is spinning at a certain speed? The next time you start up an electric motor, listen for a 'click' soon after you flip the 'on' switch, and you will understand what is happening.

So ... why does it do that?

Photo Copyright by Wayne Wooten (used with permission)
Why do spinning objects 'fly outwards'?

Most people would say that these are examples of 'centrifugal forces' ... implying that the objects swing out because there is a force pushing them away from the center. Wrong! These are examples of an objects inertia which ... according to Newton's First Law of Motion ... says that if an object is in motion, it will continue in motion ... in the same direction.

This diagram shows that if an object (shown as a red dot) is seen moving in a circle, the 'natural' motion is to move tangent to the circle (shown as a green horizontal arrow to the right). There is no mysterious force moving it away from the circle. In fact, there has to be an inward force on the object to keep it from moving along the tangent line. This force is known as the centripetal force (shown as small blue arrow pointing toward the center). In the examples above, the centripetal force is provided by ropes (amusement ride) and springs (lawnmower), and is always directed toward the center of the circle.

If we take this one step further, we need to apply Newton's Second Law of Motion. Since a net force is constantly acting on a spinning object (toward the center), it also implies that it is accelerating. Physicists call this centripetal acceleration. An acceleration is typically associated with an object moving in a straight line but going faster or slower. This is an acceleration associated with a change in direction but it is an acceleration, none the less.

If you were to tie a rope to a rock and swing it in circles over your head, YOU would be providing the centripetal force on the rock as you strain to keep the rock in its course. Since Newton showed that forces always act in pairs (Action - Reaction), the rock would pull on you with an equal but opposite force. This would be a centrifugal force and has no bearing on the behavior of the rock since we are only concerned with the force acting ON the rock ... not forces provided BY the rock.

Question #1 - Would the centripetal force change if you replaced the rock with one more massive? (answers below)

Question #2 - Would the centripetal force change if you were to spin the rock faster (more rpm's)?

More applications

Inertial switches to avoid problems in a roll-over

Inertial 'kill switch' (animation)

Often it is essential to detect and/or limit the degree of acceleration in a system. This acceleration can be linear (straight line) or centripetal (going in circles). This can be controlled by an inertial switch, or other types of on/off switches. It may be something as simple as the tiny 'swing levers' we illustrated above in the lawnmower. If the system is spinning too fast, a 'swing lever' breaks an electrical contact and the system slows down. In the case of linear accelerations and/or rotational accelerations, an inertial switch (see animation) may be employed. For example, cars are equipped with these kinds of shut-off switches so the fuel pump shuts off in case of an accident (sudden deceleration) and/or a roll-over. The inertia of the metal ball resists the changes in velocity as the device is accelerated, and its separation from the support pin causes it to trip the switch if the acceleration exceeds a certain limit. This kind of switch is much like a circuit breaker. Once it is tripped, it needs to be manually reset.

Astronauts are NOT weightless

In 1957, Russia launched the first artificial satellite to orbit the earth, Sputnik. There are several thousand crafts orbiting the earth today (most of which no longer work). Whenever you see an image from space, the astronauts floating around are usually referenced as being 'weightless' ... wrong!

Are astronauts weightless? Courtesy NASA

If you have ever taken a physics class, you learn the details about projectile motion. One of the most interesting demonstrations deals with this situation: If a person fires a bullet from a gun which is held horizontally and at the same time the gun fires, the person drops another bullet from the same height, which bullet hits the ground first? OK, I've never done this with a gun ... but using a similar setup, you will find that both bullets would hit the ground at the same time.

Projectile motion animation

The whole point is to show that the bullet fired from the gun is actually falling (due to gravity) in exactly the same way the bullet which is dropped from the hand. Of course, there are certain assumptions built into this demonstration ... this experiment is done in a vacuum (no air friction) and the ground is perfectly horizontal. Given that, the faster the projectile is fired, the farther it goes. As the projectile leaves the gun at faster and faster speeds, the bullet's path shows less and less arc ... but it still hits the ground at the same time as the other bullet because it is always falling. Keep in mind, the object is always under the influence of gravity (or it would fly off in a straight line). Click link 2.2.a for a video on this.

The faster the projectile, the less curvature to the arc (animation)

So let's suppose you were able to fire the bullet at 18,000 miles per hour? The bullet would still fall, and still have an arc, but quite a special arc. You see, the arc it would take would exactly match the curvature of the earth. Our assumption that the ground is perfectly horizontal no longer applies and the bullet simply falls ... and falls ... and falls ... and never hits the ground! We would say it is in orbit. The centripetal force comes from the gravitational attraction with the earth - i.e. weight.

Click link 2.2.band shoot a cannon ball into orbit and you should get the idea that all satellites are in a perpetual state of freefall. Now think about an astronaut inside a satellite (like the International Space Station). That person is also falling (along with the ship). Of course they appear weightless but then again, so would you if you were in an elevator and the cables broke (don't try this, please). Physics folks call this situation a 'zero g' environment which is a fancy way of saying they appear to be outside any gravitational field ... but of course, you now know that they are most definitely under the influence of gravity (and as a result, have weight). Now it is true that those same astronauts are farther from the earth's center and because gravity decreases with distance, they weigh slightly less. However, they are not weightless.

Geosynchronous orbits

To put a rocket or shuttle in orbit, all you need to do is get it out of the atmosphere, about 100 miles above the surface, so there is no friction to slow it down ... get it moving at 18,000 mph sideways ... and shut off the engines. This low altitude orbit means the payload will take about 90 minutes to orbit the earth.

If you place a satellite much farther from the earth ... just over 22,000 miles, its orbital period is one day. Place that same satellite over the equator and it appears to be 'fixed' over one location (because it now orbits with the same period as the earth's rotation). This is known as a geosynchronous orbit (or geostationary orbit). Satellites in these orbits are useful for communications and imaging (watch any weather image on TV).

Free Spinning Objects

Objects rotating about an axis (tops, flywheels, gyroscopes) are governed by laws of motion. There are many similarities between the physics of linear motion and the physics of rotational motion. Please spend a few minutes reviewing Newton's first two laws of motion presented in the last unit. You will recall that an object will tend to keep moving (or remain stationary) until a force acts on it. However, when a force does act on an object, it will accelerate. The rate of acceleration depends on the size of the force as well as the mass of the object itself. The more massive the object, the more difficult it becomes to change its state of motion. That is, you can think of 'mass' as measure of inertia.

When we adopt these laws of motion to rotation, we must consider the net torque acting on an object (not just force). Objects in rotational motion tend to keep their state of rotational motion (or remain at rest) until acted upon by a net torque. When a net torque does act on an object, you can expect it to change its rotational speed. The rate of angular acceleration depends on the size of the torque. The larger the torque, the larger the expected change in angular speed. Now let’s consider the mass (inertia) of the object. This is where you see one interesting difference between translational motion and angular motion. When we consider the rotational inertia of a spinning object, we need to consider the mass of the object as well as where that mass is distributed with respect to the axis of rotation. Physicists call this moment of inertia (I). You can think of moment of inertia as the 'rotational inertia' of an object. That is, we replace the word 'mass' (in the translational case) with 'moment of inertia' (in the rotational case). Basically, this concept is trying to tell us that a spinning object becomes more resistant to changes in motion as the mass increases and/or the mass is distributed farther from the axis of rotation.

This quantity (moment of inertia) is a bit tricky to comprehend so let me demonstrate by example. Consider two tires (at rest) of equal mass but one is mounted on a 14' rim and another on a 16' rim. The tire mounted on the 16' rim has a higher moment of inertia (since the mass is farther from the axis of rotation). As a result, it has more 'rotational inertia'. If you apply the same torque to each tire, both will spin faster, but the tire mounted on the 16' rim will undergo a lower acceleration than the tire mounted on the 14' rim. You can think of the tire mounted on the 16' rim as being more resistant to changes in rotation motion. Now suppose both tires are spinning with identical angular speeds. Do you think each tire will have the same amounts of rotational kinetic energy? No! If you guessed that the tire mounted on the 16' rim has more rotational kinetic energy, you may be understanding the concept of moment of inertia. Hint: Recall that translational KE = ½ * m * v² .. however, in the rotational case you need to consider I instead of m.

One interesting outcome of this is known as the 'ice skater effect' (the actual name is the conservation of angular momentum). It deals in instances where an object changes its moment of inertia by altering where mass is distributed on a spinning object.

The ice skater effect (animation)

To help see how this works, consider an ice skater going into a spin. The skater has a certain amount of rotational kinetic energy (which we can assume remains fairly constant once they push off). The thing that the skater changes, however, is how their body mass is distributed in space. When the skater pulls their arms in close to the body, they spin faster. This action lowers their moment of inertia and as a result, they spin faster. In order to slow down, they redistribute their mass outward, increasing their moment of inertia. Physicists have found that a quantity called angular momentum (L) is conserved (does not change) during this process. Click link 2.2.c for a demonstration.

L = I x ω

where L is angular momentum, I is moment of inertia, and ω is angular velocity.

That is, product of the moment of inertia (I) and the angular velocity (ω) will remain constant. If one quantity goes up, the other goes down to compensate. When the skater extends their arms, I increases so ω decreases.

Answer #1 - Think about Newton's Laws! Specifically, F = ma. The centripetal force increases in direct proportion to the increased mass of the rock.

Answer #2 - Common sense should answer this one. It is the reason the centrifugal governor works in the first place. Mathematically, a_r = v² / r where a_r is the centripetal acceleration, v is the linear speed and r is the radius of the circle. If the rock spins with more speed, acceleration increases and so does the force (F=ma).

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No portion may be distributed without the expressed written permission of the author

Moment Of Inertia Of Bearing

Learning Objectives

By the end of this section, you will be able to:

• Understand the relationship between force, mass and acceleration.
• Study the turning effect of force.
• Study the analogy between force and torque, mass and moment of inertia, and linear acceleration and angular acceleration.

If you have ever spun a bike wheel or pushed a merry-go-round, you know that force is needed to change angular velocity as seen in Figure 1. In fact, your intuition is reliable in predicting many of the factors that are involved. For example, we know that a door opens slowly if we push too close to its hinges. Furthermore, we know that the more massive the door, the more slowly it opens. The first example implies that the farther the force is applied from the pivot, the greater the angular acceleration; another implication is that angular acceleration is inversely proportional to mass. These relationships should seem very similar to the familiar relationships among force, mass, and acceleration embodied in Newton’s second law of motion. There are, in fact, precise rotational analogs to both force and mass.

Figure 1. Force is required to spin the bike wheel. The greater the force, the greater the angular acceleration produced. The more massive the wheel, the smaller the angular acceleration. If you push on a spoke closer to the axle, the angular acceleration will be smaller.

To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force F on a point mass m that is at a distance r from a pivot point, as shown in Figure 2. Because the force is perpendicular to r, an acceleration[latex]a=frac{F}{m}[/latex] is obtained in the direction of F. We can rearrange this equation such that F = ma and then look for ways to relate this expression to expressions for rotational quantities. We note that a = rα, and we substitute this expression into F = ma, yielding

F = mrα

Recall that torque is the turning effectiveness of a force. In this case, because F is perpendicular to r, torque is simply τ = Fr. So, if we multiply both sides of the equation above by r, we get torque on the left-hand side. That is,

rF = mr²α

or

τ = mr²α.

This last equation is the rotational analog of Newton’s second law (F = ma) where torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr² is analogous to mass (or inertia). The quantity mr² is called the rotational inertia or moment of inertia of a point mass m a distance r from the center of rotation.

Figure 2. An object is supported by a horizontal frictionless table and is attached to a pivot point by a cord that supplies centripetal force. A force F is applied to the object perpendicular to the radius r, causing it to accelerate about the pivot point. The force is kept perpendicular to r.

Making Connections: Rotational Motion Dynamics

Before we can consider the rotation of anything other than a point mass like the one in Figure 2, 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 inertiaI of an object to be the sum of mr² for all the point masses of which it is composed. That is, [latex]I=sum{text{mr}}^{2}[/latex]. Here I is analogous to m in translational motion. Because of the distance r, the moment of inertia for any object depends on the chosen axis. Actually, calculating 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 MR², where M is its total mass and R its radius. (We use M and R for an entire object to distinguish them from m and r for point masses.) In all other cases, we must consult Figure 3 (note that the table is piece of artwork that has shapes as well as formulae) for formulas for I that have been derived from integration over the continuous body. Note that I has units of mass multiplied by distance squared (kg⋅m²), as we might expect from its definition.

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

or

[latex]alpha =frac{net{tau}}{I}[/latex]

where net 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 net τ = Iα, [latex]alpha =frac{text{net}{tau}}{I}[/latex] 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.

Take-Home Experiment

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 = Iα, α = net τI, 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.

Making Connections

In statics, the net torque is zero, and there is no angular acceleration. In rotational motion, net torque is the cause of angular acceleration, exactly as in Newton’s second law of motion for rotation.

Example 1. Calculating the Effect of Mass Distribution on a Merry-Go-Round

Consider the father pushing a playground merry-go-round in Figure 4. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go-round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction.

Figure 4. A father pushes a playground merry-go-round at its edge and perpendicular to its radius to achieve maximum torque.

Strategy

Angular acceleration is given directly by the expression [latex]alpha =frac{text{net}tau}{I}[/latex]:

[latex]alpha =frac{tau }{I}[/latex]

To solve for α, we must first calculate the torque τ (which is the same in both cases) and moment of inertia I (which is greater in the second case). To find the torque, we note that the applied force is perpendicular to the radius and friction is negligible, so that

τ = rF sin θ = (1.50 m) (250 N) = 375 N ⋅ m.

Solution for (a)

The moment of inertia of a solid disk about this axis is given in Figure 3 to be

[latex]frac{1}{2}{text{MR}}^{2}[/latex],

where M = 50.0 kg and R = 1.50 m, so that

I = (0.500)(50.0 kg)(1.50 m)² = 56.25 kg⋅m².

Now, after we substitute the known values, we find the angular acceleration to be

[latex]alpha=frac{tau}{I}=frac{375 text{ N}cdottext{ m}}{56.25 text{ kg}cdottext{ m}^{2}}=6.67 frac{text{rad}}{text{s}^{2}}[/latex].

Solution for (b)

We expect the angular acceleration for the system to be less in this part, because the moment of inertia is greater when the child is on the merry-go-round. To find the total moment of inertia I, we first find the child’s moment of inertia I_c by considering the child to be equivalent to a point mass at a distance of 1.25 m from the axis. Then,

I_c = MR² = (18.0 kg)(1.25 m)² = 28.13 kg ⋅ m².

The total moment of inertia is the sum of moments of inertia of the merry-go-round and the child (about the same axis). To justify this sum to yourself, examine the definition of I:

I = 28.13 kg ⋅ m² + 56.25 kg ⋅ m² = 84.38 kg ⋅ m².

Substituting known values into the equation for α gives

[latex]alpha =frac{tau }{I}=frac{text{375 N}cdot text{m}}{text{84.38 kg}cdot {text{m}}^{2}}=4.44frac{text{rad}}{text{s}^{2}}[/latex].

Discussion

The angular acceleration is less when the child is on the merry-go-round than when the merry-go-round is empty, as expected. The angular accelerations found are quite large, partly due to the fact that friction was considered to be negligible. If, for example, the father kept pushing perpendicularly for 2.00 s, he would give the merry-go-round an angular velocity of 13.3 rad/s when it is empty but only 8.89 rad/s when the child is on it. In terms of revolutions per second, these angular velocities are 2.12 rev/s and 1.41 rev/s, respectively. The father would end up running at about 50 km/h in the first case. Summer Olympics, here he comes! Confirmation of these numbers is left as an exercise for the reader.

Check Your Understanding

Torque is the analog of force and moment of inertia is the analog of mass. Force and mass are physical quantities that depend on only one factor. For example, mass is related solely to the numbers of atoms of various types in an object. Are torque and moment of inertia similarly simple?

Solution

No. Torque depends on three factors: force magnitude, force direction, and point of application. Moment of inertia depends on both mass and its distribution relative to the axis of rotation. So, while the analogies are precise, these rotational quantities depend on more factors.

Section Summary

• The farther the force is applied from the pivot, the greater is the angular acceleration; angular acceleration is inversely proportional to mass.
• If we exert a force F on a point mass m that is at a distance r from a pivot point and because the force is perpendicular to r, an acceleration a = F/m is obtained in the direction of F. We can rearrange this equation such that

  and then look for ways to relate this expression to expressions for rotational quantities. We note that a = rα, and we substitute this expression into F = ma, yielding

  F = mrα

• Torque is the turning effectiveness of a force. In this case, because F is perpendicular to r, torque is simply τ = rF If we multiply both sides of the equation above by r, we get torque on the left-hand side. That is,

  or

  τ = mr²α.

• The moment of inertia I of an object is the sum of MR² for all the point masses of which it is composed. That is,

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

  or

  [latex]alpha =frac{text{net}tau}{I}[/latex]

Conceptual Questions

1. The moment of inertia of a long rod spun around an axis through one end perpendicular to its length is ML²/3. Why is this moment of inertia greater than it would be if you spun a point mass M at the location of the center of mass of the rod (at L/2)? (That would be ML²/4.)

2. Why is the moment of inertia of a hoop that has a mass M and a radius R greater than the moment of inertia of a disk that has the same mass and radius? Why is the moment of inertia of a spherical shell that has a mass M and a radius R greater than that of a solid sphere that has the same mass and radius?

3. Give an example in which a small force exerts a large torque. Give another example in which a large force exerts a small torque.

4. While reducing the mass of a racing bike, the greatest benefit is realized from reducing the mass of the tires and wheel rims. Why does this allow a racer to achieve greater accelerations than would an identical reduction in the mass of the bicycle’s frame?

5. A ball slides up a frictionless ramp. It is then rolled without slipping and with the same initial velocity up another frictionless ramp (with the same slope angle). In which case does it reach a greater height, and why?

Problems & Exercises

1. This problem considers additional aspects of Example 1: Calculating the Effect of Mass Distribution on a Merry-Go-Round. (a) How long does it take the father to give the merry-go-round an angular velocity of 1.50 rad/s? (b) How many revolutions must he go through to generate this velocity? (c) If he exerts a slowing force of 300 N at a radius of 1.35 m, how long would it take him to stop them?

2. Calculate the moment of inertia of a skater given the following information. (a) The 60.0-kg skater is approximated as a cylinder that has a 0.110-m radius. (b) The skater with arms extended is approximately a cylinder that is 52.5 kg, has a 0.110-m radius, and has two 0.900-m-long arms which are 3.75 kg each and extend straight out from the cylinder like rods rotated about their ends.

3. The triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 × 10³ N with an effective perpendicular lever arm of 3.00 cm, producing an angular acceleration of the forearm of 120 rad/s². What is the moment of inertia of the boxer’s forearm?

4. A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces an angular acceleration of 30.00 rad/s² and her lower leg has a moment of inertia of [latex]text{0.750 kg}cdot {text{m}}^{2}[/latex] What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90 cm?

5. Suppose you exert a force of 180 N tangential to a 0.280-m-radius 75.0-kg grindstone (a solid disk). (a)What torque is exerted? (b) What is the angular acceleration assuming negligible opposing friction? (c) What is the angular acceleration if there is an opposing frictional force of 20.0 N exerted 1.50 cm from the axis?

6. Consider the 12.0 kg motorcycle wheel shown in Figure 6. Assume it to be approximately an annular ring with an inner radius of 0.280 m and an outer radius of 0.330 m. The motorcycle is on its center stand, so that the wheel can spin freely. (a) If the drive chain exerts a force of 2200 N at a radius of 5.00 cm, what is the angular acceleration of the wheel? (b) What is the tangential acceleration of a point on the outer edge of the tire? (c) How long, starting from rest, does it take to reach an angular velocity of 80.0 rad/s?

Figure 6. A motorcycle wheel has a moment of inertia approximately that of an annular ring.

7. Zorch, an archenemy of Superman, decides to slow Earth’s rotation to once per 28.0 h by exerting an opposing force at and parallel to the equator. Superman is not immediately concerned, because he knows Zorch can only exert a force of 4.00 × 10⁷ N (a little greater than a Saturn V rocket’s thrust). How long must Zorch push with this force to accomplish his goal? (This period gives Superman time to devote to other villains.) Explicitly show how you follow the steps found in the Problem-Solving Strategy for Rotational Dynamics section (above).

8. An automobile engine can produce 200 N∙m of torque. Calculate the angular acceleration produced if 95.0% of this torque is applied to the drive shaft, axle, and rear wheels of a car, given the following information. The car is suspended so that the wheels can turn freely. Each wheel acts like a 15.0 kg disk that has a 0.180 m radius. The walls of each tire act like a 2.00-kg annular ring that has inside radius of 0.180 m and outside radius of 0.320 m. The tread of each tire acts like a 10.0-kg hoop of radius 0.330 m. The 14.0-kg axle acts like a rod that has a 2.00-cm radius. The 30.0-kg drive shaft acts like a rod that has a 3.20-cm radius.

9. Starting with the formula for the moment of inertia of a rod rotated around an axis through one end perpendicular to its length [latex]left(I={mathrm{Mell }}^{ 2}/3right)[/latex], prove that the moment of inertia of a rod rotated about an axis through its center perpendicular to its length is [latex]I={mathrm{Mell }}^{ 2}/12[/latex]. You will find the graphics in Figure 3 useful in visualizing these rotations.

10. Unreasonable Results A gymnast doing a forward flip lands on the mat and exerts a 500-N ∙ m torque to slow and then reverse her angular velocity. Her initial angular velocity is 10.0 rad/s, and her moment of inertia is [latex]0.050text{kg}cdot {text{m}}^{2}[/latex]. (a) What time is required for her to exactly reverse her spin? (b) What is unreasonable about the result? (c) Which premises are unreasonable or inconsistent?

11. Unreasonable Results An advertisement claims that an 800-kg car is aided by its 20.0-kg flywheel, which can accelerate the car from rest to a speed of 30.0 m/s. The flywheel is a disk with a 0.150-m radius. (a) Calculate the angular velocity the flywheel must have if 95.0% of its rotational energy is used to get the car up to speed. (b) What is unreasonable about the result? (c) Which premise is unreasonable or which premises are inconsistent?

Glossary

torque:

the turning effectiveness of a force

rotational inertia:

resistance to change of rotation. The more rotational inertia an object has, the harder it is to rotate

moment of inertia:

mass times the square of perpendicular distance from the rotation axis; for a point mass, it isI = mr² and, because any object can be built up from a collection of point masses, this relationship is the basis for all other moments of inertia

Selected Solutions to Problems & Exercises

1. (a) 0.338 s (b) 0.0403 rev(c) 0.313 s

3. [latex]text{0.50 kg}cdot {text{m}}^{2}[/latex]

5. (a) [latex]50.4 Ncdot text{m}[/latex] (b) 17.1 rad/s² (c) 17.o rad/s²

7. 3.96 × 10¹⁸ s or 1.26 × 10¹¹ y

9. [latex]begin{array}{c}{I}_{text{end}}={I}_{text{center}}+m{left(frac{l}{2}right)}^{2} text{Thus,}{I}_{text{center}}={I}_{text{end}}-frac{1}{4}{text{ml}}^{2}=frac{1}{3}{text{ml}}^{2}-frac{1}{4}{text{ml}}^{2}=frac{1}{text{12}}{text{ml}}^{2}end{array}[/latex]

10. (a) 2.0 ms (b) The time interval is too short. (c) The moment of inertia is much too small, by one to two orders of magnitude. A torque of [latex]text{500 N}cdot text{m}[/latex] is reasonable.

11. (a) 17,500 rpm (b) This angular velocity is very high for a disk of this size and mass. The radial acceleration at the edge of the disk is > 50,000 gs. (c) Flywheel mass and radius should both be much greater, allowing for a lower spin rate (angular velocity).