Topic 6 – Rotation
Chapter 9
I. 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.
II. 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 (Figure9-2), 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 Figure9-2, 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 2ri.
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 , 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 .
i. The linear velocity 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 .
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 , 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 / UnitsPosition / x / m / Angle / / rad
Displacement / s / m / Angular Displacement / / rad
Velocity (initial) / vo / m/s / Angular Velocity (Initial) / o / rad/s
Velocity (final) / v1 / m/s / Angular Velocity (Final) / 1 / rad/s
Acceleration / a / m/s2 / Angular Acceleration / / 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 inRADIANS!
III. 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 Figure9-3. 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 Equation9-11 and Equation9-12. (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 Figure9-4.
e. Special Notes on Section 9 – 2.
IV. 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 (Figure9-7). 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 (2pr) times its differential width (Rdq).
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 (Figure9-10).
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 (Figure9-11) 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 Figure9-13, 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?