Rotational motion is all around us



Introduction.

a. Rotational motion is all around us. Earth rotates about its axis. Wheels, gears, propellers, motors, the drive shaft in a car, a CD in its player, a pirouetting ice skater, all rotate.

b. In this chapter, we consider rotation about an axis that is fixed in space, as in a merry-go-round, or about an axis that is moving without changing its direction in space, as in a rolling wheel on a car that is traveling in a straight line. The study of rotational motion is continued in Chapter 10 where more general examples of rotational motion are considered.

I. Rotational Kinematics: Angular Velocity and Angular Acceleration.

a. Every point of a rigid object rotating about a fixed axis moves in a circle whose center is on the axis and whose radius is the radial distance from the axis of rotation to that point. A radius drawn from the rotation axis to any point on the body sweeps out the same angle in the same time.

b. Let ri be the distance from the center of the disk to the ith particle (Figure 9-2), and let θi be the angle measured counterclockwise from a fixed reference line in space to a radial line from the axis to the particle. As the disk rotates through an angle dθ, the particle moves through a circular arc of directed length dsi.

i. If counterclockwise is designated as the positive direction, then dθ, θi, and dsi, shown in Figure 9-2, are all positive.

ii. If clockwise is designated the positive direction, they are all negative.

iii. The angle θi, the directed length dsi, and the distance ri vary from particle to particle, but the ratio dsi/ri, called the angular displacement dθ, is the same for all particles of the disk.

iv. For one complete revolution, the arc length si is 2[pic]ri.

c. The time rate of change of the angle is the same for all particles of the disk, and is called the angular velocity ω of the disk.

i. The instantaneous angular velocity ω is an angular displacement of short duration divided by the time.

ii. All points on the disk undergo the same angular displacement during the same time, so they all have the same angular velocity.

iii. The SI units of ω are rad/s. Because radians are dimensionless, the dimension of angular velocity is that of reciprocal time, T−1.

iv. The magnitude of the angular velocity is called the angular speed.

v. Practice Problem 9 – 1: A compact disk is rotating at 3000 rev/min. What is its angular speed in radians per second?

d. Angular acceleration is the rate of change of angular velocity.

i. If the rotation rate of a rotating object increases, the angular speed |ω| increases.

ii. If |ω| is increasing, and if the angular velocity ω is clockwise, then the change in the angular velocity Δω is also clockwise.

iii. If the rotation rate decreases, then both Δω and αav are in the opposite direction to ω.

e. The instantaneous rate of change of angular velocity is called the angular acceleration α.

i. The SI units of α are rad/s2. α is positive if ω is increasing, and α is negative if ω is decreasing.

f. The angular displacement θ, the angular velocity ω, and angular acceleration α are analogous to the linear displacement x, linear velocity [pic], and linear acceleration ax in one-dimensional motion.

g. Example 9 – 1: A compact disk rotates from rest to 500 rev/min in 5.5 s. (a) What is its angular acceleration, assuming that it is constant? (b) How many revolutions does the disk make in 5.5 s? (c) How far does a point on the rim 6.0 cm from the center of the disk travel during the 5.5 s it takes to get to 500 rev/min?

h. Practice Problem 9 – 2: (a) Convert 500 rev/min to rad/s. (b) Check the result of Part (b) in the example using [pic].

i. The linear velocity [pic]of a particle on the disk is tangent to the circular path of the particle and has magnitude dsi/dt. We can relate this tangential velocity to the angular velocity of the disk.

j. The tangential acceleration of a particle on the disk is [pic].

k. Each particle of the disk also has a centripetal acceleration, which points inward along the radial line.

i. Practice Problem 9 – 3: A point on the rim of a compact disk is 6.00 cm from the axis of rotation. Find the tangential speed [pic], tangential acceleration at, and centripetal acceleration ac of the point when the disk is rotating at a constant angular speed of 300 rev/min.

ii. Practice Problem 9 – 4: Find the linear speed of a point on the CD in Example 9 – 1 at (a) r = 2.40 cm, when the disk rotates at 500 rev/min, and (b) r = 6.00 cm, when the disk rotates at 200 rev/min.

Where:

|Variable |Symbol |Units |Rot  Variable |Symbol |Units |

|Displacement |s |m |Angular Displacement |[pic] |rad |

|Velocity (initial) |vo |m/s |Angular Velocity (Initial) |[pic]o |rad/s |

|Velocity (final) |v1 |m/s |Angular Velocity (Final) |[pic]1 |rad/s |

|Acceleration |a |m/s2 |Angular Acceleration |[pic] |rad/s2 |

|Time |t |s |Time |t |s |

|Force (Total) |F |N |Torque |T |Nm |

|Mass |M |kg |Mass Moment of Inertia |I |kgm2 |

Note: These formulas only work in RADIANS!

II. Rotational Kinetic Energy.

a. The kinetic energy of a rigid object rotating about a fixed axis is the sum of the kinetic energies of the individual particles that collectively constitute the object.

b. The sum in the expression farthest to the right is the object’s moment of inertia I for the axis of rotation.

c. Example 9 – 2: An object consists of four point particles, each of mass m, connected by rigid massless rods to form a rectangle of edge lengths 2a and 2b, as shown in Figure 9-3. The system rotates with angular speed ω about an axis in the plane of the figure through the center, as shown. (a) Find the kinetic energy of this object using Equation 9-11 and Equation 9-12. (b) Check your result by individually calculating the kinetic energy of each particle and then taking their sum.

d. Practice Problem 9 – 5: Find the moment of inertia of this system for rotation about an axis parallel to the first axis but passing through two of the particles, as shown in Figure 9-4.

e. Special Notes on Section 9 – 2.

III. Calculating the Moment of Inertia.

a. The moment of inertia about an axis is a measure of the inertial resistance of the object to changes in its rotational motion about the axis. It is the rotational analog of mass.

b. Unlike the mass of an object, which is a property of the object itself, the moment of inertia depends on the location of the axis as well as the mass distribution of the object.

c. For systems consisting of discrete particles, we can compute the moment of inertia about a given axis.

i. Example 9 – 3: Estimating Moment of Inertia - Estimate the moment of inertia of a thin uniform rod of length L and mass M about an axis perpendicular to the rod and through one end. Execute this estimation by modeling the rod as three point masses, each point mass representing one-third of the rod.

ii. Practice Problem 9 – 6: The contribution to the moment of inertia of the third of the rod farthest from the axis is many times greater than is the contribution of the third closest to the axis. About how many times greater is it?

d. To calculate the moment of inertia for continuous objects, we imagine the object to consist of a continuum of very small mass elements.

where r is the radial distance from the axis to mass element dm. To evaluate this integral, we first express dm as a density times an element of length, area, or volume.

e. We can calculate I for continuous objects of various shapes.

Example 9 – 4: Moment of Inertia of a Thin Uniform Rod -Find the moment of inertia of a thin uniform rod of length L and mass M about an axis perpendicular to the rod and through one end.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Hoop about a perpendicular axis through its center Assume that a hoop has mass M and radius R (Figure 9-7). The axis of rotation is the symmetry axis of the hoop, which is perpendicular to the plane of the hoop and through its center.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform disk about a perpendicular axis through its center

The uniform disk is made of many hoops. The area of the disk has a constant R, but the radius of each hoop varies and is differential. The hoop radius = r.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform solid cylinder about its axis We consider a cylinder to be a set of disks. The radius of each disk is constant and is equal to R.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform hollow cylinder shell about its axis We consider a hollow cylinder shell to be a set of hoops. Each hoop has a constant radius, R. All the mass is the same distance from the center.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform solid cylinder about its axis. We consider a sd cylinder to be a set of cylinder shells. The radius of each shell varies and is differential.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform washer about its axis. We consider the washer to be made of smaller hoops, each with a different radius. Therefore the radius is differential.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform hollow cylinder about its axis. We consider the hollow cylinder to be made of washers, each with a constant radius.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia:

Uniform Hollow Spherical Shell about its axis. We can think of the hollow spherical shell to be made of many differential rings. Each ring has a different radius. The differential area of this ring is its circumference (2πr) times its differential width (Rdθ).

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia

Uniform Solid Sphere about its axis. We can think of the solid sphere as being made of many spherical shells, each with a different radius. The radius is differential.

Length Area Volume

Differential L, A, or V

Density:

Differential Mass:

Moment of Inertia

f. We can often simplify the calculation of moments of inertia for various objects by using the parallel-axis theorem, which relates the moment of inertia about an axis through the center of mass to the moment of inertia about a second, parallel axis (Figure 9-10).

i. Let I be the moment of inertia, and let I cm be the moment of inertia about a parallel axis through the center of mass.

ii. In addition, let M be the total mass of the object and let h be the distance between the two axes.

iii. Example 9 – 5: Applying the Parallel Axis Theorem - A thin uniform rod of mass M and length L on the x axis (Figure 9-11) has one end at the origin. Using the parallel-axis theorem, find the moment of inertia about the y’ axis, which is parallel to the y axis, and through the center of the rod.

iv. Practice Problem 9 – 7: Using the parallel-axis theorem, show that when comparing the moments of inertia of an object about two parallel axes, the moment of inertia is less about the axis that is nearer to the center of mass.

v. Example 9 – 6: A Flywheel – Powered Car - You are driving an experimental hybrid vehicle that is designed for use in stop-and-go traffic. In a car with conventional brakes, each time you brake to a stop the kinetic energy is dissipated as heat. In this hybrid vehicle, the braking mechanism transforms the translational kinetic energy of the vehicle’s motion into the rotational kinetic energy of a massive flywheel. As the car returns to cruising speed, this energy is transferred back into the translational kinetic energy of the car. The 100-kg flywheel is a hollow cylinder with an inner radius R1 of 25.0 cm, an outer radius R2 of 40.0 cm, and a maximum angular speed of 30,000 rev/min. On a dark and dreary night, the car runs out of gas 15.0 mi from home with the flywheel spinning at maximum speed. Is there sufficient energy stored in the flywheel for you and your nervous grandmother to make it home? (When driving at the minimum highway speed of 40.0 mi/h, air drag and rolling friction dissipate energy at 10.0 kW.)

vi. Example 9 – 7: The Pivoted Rod - A uniform thin rod of length L and mass M, pivoted at one end as shown in Figure 9-13, is held horizontal and then released from rest. Assuming that effects due to friction and air resistance are negligible, find (a) the angular speed of the rod as it sweeps through the vertical position, and (b) the force exerted on the rod by the pivot at this instant. (c) What initial angular speed would be needed for the rod to just reach a vertical position at the top of its swing?

vii. Example 9 – 8: A Winch and a Bucket - A winch is at the top of a deep well. The drum of the winch has mass mw and radius R. Virtually all its mass is concentrated a distance R from the axis. A cable wound around the drum suspends a bucket of water of mass mb. The entire cable has mass mc and length L. Just when you have the bucket at the highest point, your hand slips and the bucket falls back down the well, unwinding the winch cable as it goes. How fast is the bucket moving after it has fallen a distance d, where d is less than L? Assume that effects due to friction and air resistance are negligible.

IV. Newton’s Second Law for Rotation.

a. To set a top spinning, you twist it. In Figure 9-17, a disk is set spinning by the forces [pic]and [pic]exerted at the edges of the disk in the tangential direction.

b. The directions of these forces and their points of application are important. If the same forces are applied at the same points but in the radial direction (Figure 9-18a), the disk will not start to spin. In addition, if the same forces are applied in the tangential direction, but at points closer to the center of the disk (Figure 9-18b), the disk will not gain angular speed as quickly.

c. Figure 9-19 shows a particle of mass m attached to one end of a massless rigid rod of length r. The rod is free to rotate about a fixed axis perpendicular to the rod and passing through the end of the rod at A. Consequently, the particle is constrained to move in a circle of radius r. A single force [pic]is applied to the particle as shown.

d. Applying Newton’s second law to the particle and taking components in the tangential direction gives

e. The product rFt is the torque [pic]about the rotation axis associated with the force.

f. Torque about a point is defined as a vector quantity in Chapter 10. What we refer to as “torque about an axis” is the component of the torque vector parallel with the axis.

g. In Chapter 8, we saw that the net force acting on a system of particles is equal to the net external force acting on the system because the internal forces (those exerted by the particles within the system on one another) cancel in pairs. The treatment of internal torques exerted by the particles within a system on one another leads to a similar result, that is, the net torque acting on a system equals the net external torque acting on the system.

h. Figure 9-20 shows a force [pic]acting on an object constrained to rotate about a fixed axis A, not shown, which passes through O and is perpendicular to the page.

i. The positive tangential direction is shown at the point of application of the force, and r is the radial distance of this point of application from axis A.

ii. The torque [pic]due to this force about axis A is [pic]. In principle, the expression Ftr is all that is needed to calculate torques.

iii. However, in practice, calculations are often simpler if alternative expressions for torque are used.

iv. The line of action of a force is the line through the point of application of the force that is parallel to the force.

v. The moment arm [pic]is the perpendicular distance between A and the line of action. (The moment arm is also called the lever arm.)

vi. The torque of a force about an axis is also called the moment of the force about the axis.

i. The net gravitational torque can be calculated by considering the total gravitational force (the sum of the microscopic gravitational forces) to act at a single point-the center of gravity.

i. The torque due to a uniform gravitational field is calculated as if the entire gravitational force is applied at the center of gravity.

ii. For any object in a uniform gravitational field, the center of gravity and the center of mass coincide.

V. Applications of Newton’s Second Law for Rotation.

a. Applying Newton’s second law for Rotation

i. PICTURE Angular accelerations for rigid objects can be found by using free-body diagrams and Newton’s second law for rotation, which is [pic]. If [pic]is constant, then the constant angular acceleration equations apply. Time intervals and angular positions, velocities, and angular accelerations can then be determined using these equations.

ii. SOLVE

1. Draw a free-body diagram with the object shown as a likeness of the object (not just as a dot).

2. Draw each force vector along the line of action of that force.

3. On the diagram indicate the positive direction (clockwise or counterclockwise) for rotations.

iii. CHECK Make sure that the signs of your results are consistent with your choice for the positive directions of rotation.

b. Problems.

i. Example 9 – 9: A Stationary Bike - To get some exercise without going anywhere, you set your bike on a stand so that the rear wheel is free to turn. As you pedal, the chain applies a force of 18 N to the rear sprocket wheel at a distance of rs = 7.0 cm from the rotation axis of the wheel. Consider the wheel to be a hoop (I = MR2) of radius R = 35 cm and mass M = 2.4 kg. What is the angular velocity of the wheel after 5.0 s?

ii. Example 9 – 10: A Uniform Rod, Pivoted at One End - A uniform thin rod of length L and mass M is pivoted at one end. It is held horizontal and released. Effects due to friction and air resistance are negligible. Find (a) the angular acceleration of the rod immediately following its release, and (b) the magnitude of the force FA exerted on the rod by the pivot at this instant.

iii. Practice Problem 9 – 8: A small pebble of mass [pic]is placed on top of the rod at its center. Just after the rod is released find (a) the acceleration of the pebble, and (b) the force it exerts on the rod.

c. Nonslip Conditions.

i. In physics courses, there are many situations in which a taut string is in contact with a rotating pulley wheel. For the string not to slip on the pulley wheel, the parts of the string and the wheel that are in direct contact with each other must share the same tangential velocity.

where [pic]is the tangential velocity of the string and Rω is the tangential velocity of the perimeter of the pulley wheel. The wheel has radius R and is rotating with angular velocity ω.

ii. Differentiating both sides of the nonslip condition with respect to time gives

where at is the tangential acceleration of the string and α is the angular acceleration of the wheel.

iii. Example 9 – 11: Tension in a String - An object of mass m is suspended from a light string that is wound around the rim of a pulley wheel that has moment of inertia I and radius R. The wheel bearing is frictionless and the string does not slip on the rim. The wheel is released from rest. It starts to rotate as the object descends and the string unwinds. Find the tension in the string and the acceleration of the object.

iv. Example 9 – 12: Two Block and a Pulley I - The system shown in Figure 9-28 is released from rest. The mass of the pulley wheel is not negligible, but the friction in the bearing is negligible. The string does not slip on the pulley wheel. Given that m1 > m2 what can be determined about the tensions T1 and T2?

v. Example 9 – 13: Two Blocks and a Pulley II - Two blocks are connected by a string that passes over a pulley of radius R and moment of inertia I. The block of mass m1 slides on a frictionless, horizontal surface; the block of mass m2 is suspended from the string (Figure 9-29). Find the acceleration a of the blocks and the tensions T1 and T2. The string does not slip on the pulley.

vi. Power.

1. When you spin up an object, you do work on it, increasing its kinetic energy.

a. Consider the force [pic]acting on a rotating object.

b. As the object rotates through an angle dθ, the point of application of the force moves a distance ds = r dθ, and the force does work.

c. The rate at which the torque does work is the power input of the torque.

d. Equations in b and c are analogous to

2. Example 9 – 14: Torque Exerted by an Automobile Engine - The maximum torque produced by the 5.4-L V8 engine of a 2005 Ford GT is 678 N · m of torque at 4500 rev/min. Find the power output of the engine operating at these maximum torque conditions.

3. Practice Problem 9 – 9: The maximum power produced by the Ford GT engine is 500 hp at 6000 rev/min. What is the torque when the engine is operating at maximum horsepower?

4. Example 9 – 15: Stopping the Wheel - The specifications for the London Eye include that it be able to brake to a stop so that the passenger compartments move no more than 10 m during braking. The operating speed of the 135-m-diameter 1600-tonne wheel is 2.0 rev/h. (One tonne equals 1000 kg.) A picture of the wheel can be found at the beginning of this chapter. (a) Estimate the torque that is required to stop the wheel so the rim travels 10 m during the braking. (b) Assuming that the braking force is applied at the rim, what is the magnitude of the breaking force?

VI. Rolling Objects.

a. Rolling Without Slipping.

i. When a spool rolls without slipping down an incline (Figure 9-31), the points of the spool in contact with the incline are instantaneously at rest and the spool rotates about a rotation axis through the contact points.

ii. Point P on the wheel moves as shown where r is the radial distance from the rotation axis to point P.

iii. For a point on the very top of the wheel, r = 2R, so the top of the wheel is moving at twice the speed of the center of the wheel.

iv. A wheel of radius R is rolling without slipping along a straight path. As the wheel rotates through angle [pic](Figure 9-33), the point of contact between the wheel and the surface moves a distance s that is related to [pic].

v. If the wheel is rolling on a flat surface, the wheel’s center of mass remains directly over the point of contact, so it also moves through distance R[pic].

vi. In Chapter 8 we saw (Equation 8-7) that the kinetic energy of a system can be written as the sum of its translational kinetic energy [pic]and the kinetic energy relative to the center of mass Krel. For an object that is rotating, the kinetic energy relative to an inertial frame moving with center of mass [pic].

vii. Remember, a rolling object has both translational and rotational kinetic energy.

viii. Example 9 – 16: A Bowling Ball - A bowling ball that has an 11-cm radius and a 7.2-kg mass is rolling without slipping at 2.0 m/s on a horizontal ball return. It continues to roll without slipping up a hill to a height h before momentarily coming to rest and then rolling back down the hill. Model the bowling ball as a uniform sphere and find h.

ix. PRACTICE PROBLEM 9-10 Find the initial kinetic energy of the ball.

x. Example 9 – 17: Playing Pool - A cue stick strikes a cue ball horizontally at a point a distance d above the center of the ball (Figure 9-35). Find the value of d for which the cue ball will roll without slipping from the beginning. Express your answer in terms of the radius R of the ball.

xi. When an object rolls down an incline, its center of mass accelerates. The analysis of such a problem is simplified by an important theorem concerning the center of mass:

xii. Equation 9-31 is the same as Equation 9-19, except that in Equation 9-31 the torques and the moment of inertia are computed from a reference frame moving with the center of mass.

1. When the center of mass is accelerating (a ball rolling down an incline, for example), the center-of-mass reference frame is a noninertial one, where we would not expect our equations for Newton’s second law for rotation to be valid.

2. Nevertheless, they are valid for this special case.

xiii. Example 9 – 19: Acceleration of a Ball That Is Rolling Without Slipping – A uniform solid ball of mass m and radius R rolls without slipping down a plane inclined at an angle [pic]above the horizontal. Find the frictional force and the acceleration of the center of mass.

1. The results of steps 5 and 6 in Example 9-18 apply to any round object with the center of mass at the geometric center that is rolling without slipping.

2. For such objects, [pic]for a solid sphere, [pic]for a rolling solid cylinder, 1 for a thin hollow cylinder, and so forth. (These values of β are obtained from the expressions for I found in Table 9-1.)

xiv. The linear acceleration of any object rolling without slipping down an incline is less than g sin [pic]because of the frictional force directed up the incline.

1. Note that these accelerations are independent of the mass and the radius of the objects.

2. That is, all uniform solid spheres will roll without slipping down an incline with the same acceleration.

3. However, if we release a sphere, a cylinder, and a hoop at the top of an incline, and if they all roll without slipping, the sphere will reach the bottom first because it has the greatest acceleration.

xv. Static frictional forces do no work on the rolling objects, and if there is no slipping there is no dissipation of energy.

1. Therefore, we use the conservation of mechanical energy to find the final speed of an object released from rest that is rolling without slipping down an incline.

2. At the top of the incline, the total energy is the potential energy mgh. At the bottom, the total energy is kinetic energy.

3. For a cylinder, with [pic]. Note that this speed is independent of both the mass and the radius of the cylinder.

4. For an object rolling without slipping down an incline, the frictional force fs is less than or equal to its maximum value.

5. Practice Problem 9 – 11: A uniform cylinder rolls down a plane inclined at [pic]= 50°. What is the minimum value of the coefficient of static friction for which the cylinder will roll without slipping?

6. Practice Problem 9 – 12: For a uniform hoop of mass m that is rolling without slipping down an incline, (a) what is the force of friction, and (b) what is the maximum value of tan [pic]for which the hoop will roll without slipping?

b. Rolling With Slipping.

i. When an object slips (skids) as it rolls, the nonslip condition [pic]does not hold.

ii. The kinetic frictional force will both reduce its linear speed [pic](Figure 9-39) and increase its angular speed ω until the nonslip condition [pic]is reached, after which the ball rolls without slipping.

iii. Another example of rolling with slipping is a ball with topspin, such as a cue ball struck at a point higher than [pic]above the center (see Example 9-17) so that [pic]. Then the kinetic frictional force both increases [pic]and decreases ω until the nonslip condition [pic]is reached (Figure 9-40).

iv. Example 9 – 19: A Skidding Bowling Ball - A bowling ball of mass M and radius R is released at floor level so that at release it is moving horizontally with speed [pic]and is not rotating. The coefficient of kinetic friction between the ball and the floor is μk = 0.080. Find (a) the time the ball skids along the floor (after which it begins rolling without slipping), and (b) the distance the ball skids.

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