Rotational Mechanics
Contents 1 Introduction
2
2 Rotational Kinematics 2.1 Angular displacement, velocity, & acceleration . . . . . . . . . . . . . . . . . . . . . 2.2 Rotation with constant angular acceleration . . . . . . . . . . . . . . . . . . . . . . 2.3 Relationship between linear and angular quantities . . . . . . . . . . . . . . . . . . .
2 2 4 4
3 Rotational Energy
6
4 Calculation of Moment of Inertia
7
5 Rotational Dynamics 5.1 Torque . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.2 Relation between Torque and Angular acceleration . . . . . . . . . . . . . . . . . . .
9 9 11
6 Work, Power and Energy 6.1 Work and power . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Work and energy in rotational motion . . . . . . . . . . . . . . . . . . . . . . . . .
14 14 14
1
2
Rotational Kinematics
1
Introduction
When an extended object, such as a wheel, rotates about its axis, the motion cannot be analyzed by treating the object as a particle because at any given time different parts of the object have different linear velocities and linear accelerations. For this reason, it is convenient to consider an extended object as a large number of particles, each of which has its own linear velocity and linear acceleration. In dealing with a rotating object, analysis is greatly simplified by assuming that the object is rigid. A rigid body is one that is non–deformable, that is, it is an object in which the separations between all pairs of particles remain constant. All real bodies are deformable to some extent; however, our rigid–body model is useful in many situations in which deformation is negligible. We begin by stating without any proof, the following theorem concerned with the general motion of a rigid body: ⊙ Theorem 1 (Chasle’s Theorem). The most general displacement of a rigid body is a translation plus a rotation. We already know how to describe a translational motion1 . We now turn to description of rotational motion about a fixed axis, which is also referred to as pure rotational motion.
2
Rotational Kinematics
Definition 1 (Rotational motion). If a body moves so that along some straight line, all the particles of the body (or a hypothetical extension of the body) have zero velocity relative to some reference frame, the body is said to be in rotation relative to this reference frame. The line of stationary particles is called the axis of rotation. The feature characterizing rotational motion is that, every particle of the rigid body moves in a circular path (different radii for different particles) whose centers lie on the axis of rotation.
2.1
Angular displacement, velocity, & acceleration
Consider a rigid body rotating about an axis that is fixed in a given reference frame (see Figure 1). Suppose it accomplishes an infinitesimal rotation during the time interval dt. We shall → describe the corresponding angular displacement by the vector d− φ whose magnitude is equal to the rotation angle and whose direction coincides with the axis OO′ , with the rotation direction → obeying the right–hand thumb rule with respect to direction of vector d− φ (Figure 2). Now let us find the elementary linear displacement of any point A of the rigid body resulting → from such a rotation. The location of the point A is specified by the position vector − r drawn from a certain point O on the axis (O is thus the origin). Then the linear displacement of the end → point of the position vector − r is associated with the rotation angle dφ by the relation (Figure 1): → |d− r | = r sin θ dφ, or in vector form by the vector product → → → d− r = d− φ ×− r.
(1)
1
In this kind of motion, any straight line fixed to rigid body remains parallel to its initial orientation. Translation can be described exactly like particle kinematics.
Anant Kumar
Mob. No. 8967881837, 9932347531
3
Rotational Kinematics O′ ω ⃗
dφ
d⃗ φ
ω
A
d⃗r
⃗r
θ
O
ω
Figure 1: A rigid body rotating about a fixed axis.
Figure 2: The right–hand thumb rule for determining the direction of angular displacement, velocity, and acceleration.
→ Note that this equality holds only for an infinitesimal rotation d− φ . In other words, only in→ 2 finitesimal rotations can be treated as vectors . Moreover, the vector introduced (d− φ) can be shown to satisfy the basic property of vectors, that is, vector addition. → → → Note that in treating such quantities as position vector − r , velocity − v , acceleration − a we did not hesitate over the choice of their direction: it naturally followed from the properties of the quantities themselves. Such vectors are called polar vectors. On the other hand, such vectors as → d− φ , whose direction is specified by the rotation direction are called axial. Now let us introduce the concept of angular velocity and angular acceleration. The angular → velocity vector − ω is defined as the time rate at which the angular displacement changes: → d− φ − → , ω ≡ dt
(2)
→ where dt is the time interval during which a body performs the rotation given by d− φ . The vector − → − → ω is axial and its direction coincides with that of the vector d φ . → The time variation of the vector − ω is defined by the angular acceleration: → d− ω − → α ≡ . dt
(3)
→ → The direction of the vector − α coincides with the direction of the infinitesimal increment d− ω of − → − → − → the vector ω . Both α and ω are axial. The representation of angular velocity and angular acceleration in vector form proves to be very beneficial, especially in the study of more complicated kind of motion of a rigid body. 2
In case of finite rotation through an angle δφ, the linear displacement of the point A can be found from Figure 1 ) ( ∆φ − → . |∆ r | = r sin θ · 2 sin 2 → Whence, it is immediately seen that the displacement ∆− r cannot be represented as a vector cross product.
as:
Anant Kumar
Mob. No. 8967881837, 9932347531
4
Rotational Kinematics
Finally, we write the expressions for angular velocity and angular acceleration via projections on the rotation axis z whose positive direction is associated with the positive direction of the coordinate φ, the rotation angle, in accordance with the right–hand thumb rule. Then the → → projections ωz and αz of the vectors − ω and − α on the z axis are defined by the following formulae: dφ , dt dωz αz = . dt ωz =
(4) (5)
Here, ωz and αz are algebraic quantities. Their sign specifies the direction of the corresponding → vector. For example, if ωz > 0, then the direction of the vector − ω coincides with the positive − → direction of the z axis; and if ωz < 0, then the vector ω has the opposite direction. The same is true for angular acceleration. Example 1. A solid rotates about a stationary axis in accordance with the law φ = at − bt2 /2, where a and b are positive constants. Let us determine the motion characteristics of this body. In accordance with (4) and (5) ωz = a − bt;
αz = −b = constant.
Whence it is seen that the body performs a uniformly decelerated rotation (αz < 0), comes to a standstill at the instant t0 = a/b and then reverses its rotation direction (due to ωz changing its sign to the opposite.)
2.2
Rotation with constant angular acceleration
In our study of linear motion, we found that the simplest form of accelerated motion to analyze is motion under constant linear acceleration. Likewise, for rotational motion about a fixed axis, the simplest accelerated motion to analyze is motion under constant angular acceleration. If we write (5) in the form dωz = αz dt, and let the initial time ti = 0 and the final time tf = t, we can integrate directly to obtain ωz = ωz0 + αz t,
(for constant αz )
(6)
where ωz0 is the z component of angular velocity at the initial time. Integrating this last expression again, and taking into account that at the initial moment the angular position was φ0 , we obtain 1 (for constant αz ) (7) φ = φ0 + ωz t + αz t2 . 2 Finally, if we eliminate t from these last two equations, we obtain 2 ωz2 = ωz0 + 2αz (φ − φ0 ).
(for constant αz )
(8)
Notice that these kinematic expressions for rotational motion under constant angular acceleration are of the same form as those for linear motion under constant linear acceleration with the substitutions x → φ, v → ωz , and a → αz .
2.3
Relationship between linear and angular quantities
→ Let us find the velocity − v of an arbitrary point A of a rigid body rotating about a stationary → ′ axis OO at an angular velocity − ω (see Figure 3). Let the location of the point A be specified − → by the position vector r measured relative to the point O on the axis of rotation. Dividing both sides of Equation (1) by the corresponding time interval dt and taking into account that → → → → d− r /dt = − v and d− φ /dt = − ω , we obtain − → → → v =− ω ×− r, Anant Kumar
(9) Mob. No. 8967881837, 9932347531
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Rotational Kinematics ′
O
⃗ ω
⃗v
ωdt ρ
A
d⃗r
⃗r θ
O
Figure 3: As a rigid object rotates about the fixed axis OO′ , the point A has a linear velocity that is always tangent to the circular path of radius ρ.
→ that is, the linear velocity − v of a point A of a rigid body rotating about some axis at an angular − → → → velocity ω is equal to the vector cross product of − ω and the position vector − r of the point A relative to some point on the axis of rotation (Figure 3). → The magnitude of the vector − v is given by v = ωr sin θ, or as seen from Figure 3, v = ωρ, where ρ is the radius of the circle which the point A circumscribes. → Next, we differentiate (9) to obtain the linear acceleration − a of the point A: → ) d− v d (− → → − → = ω ×− r a = dt dt ( ) ( ) → → d− ω d− r → → ×− r + − ω × = dt dt − → − → − → − → = α × r + ω × v (→ − ) → → → =− α ×− r +− ω × − ω ×→ r Therefore, the linear acceleration is (→ − ) − → → → → a =− α ×− r +− ω × − ω ×→ r .
(10)
→ → In this case (when the rotation axis is stationary), − α is parallel to − ω (and therefore the vector ) − → − → − → − → − → − → α × r represents the tangential acceleration a t . The vector ω × ω × r is the normal → acceleration − a n . The magnitudes of these two vectors are at = αρ ;
an = ω 2 ρ,
whence the magnitude of the total acceleration is √ √ a = a2t + a2n = ρ α2 + ω 4 . Anant Kumar
Mob. No. 8967881837, 9932347531
6
Rotational Energy
3
Rotational Energy
Let us now look at the kinetic energy of a rotating rigid object, considering the object as a → collection of particles and assuming it rotates about a fixed OO′ axis with angular velocity − ω (Figure 3). Each particle has kinetic energy determined by its mass and linear speed. If the mass → of the ith particle is mi and its linear velocity is − vi , its kinetic energy is 1 mi vi2 . 2
Ki =
To proceed further, we must recall that although every particle in the rigid object has the same angular speed ω, the individual linear speeds depend on the distance ρi from the axis of rotation according to the expression vi = ρi ω. The total kinetic energy of the rotating rigid object is the sum of the kinetic energies of the individual particles: KR =
∑
Ki =
∑1 i
i
2
mi vi2 =
1∑ mi ρ2i ω 2 . 2 i
We can write this expression in the form 1 KR = 2
( ∑
) mi ρ2i
ω2.
(11)
i
where we have factored ω 2 from the sum because it is common to every particle. We simplify this expression by defining the quantity in parentheses as the moment of inertia I of the rigid body about the axis OO′ : ∑ I≡ mi ρ2i . (12) i
From the definition of moment of inertia, we see that it has dimensions of M L2 (kg·m2 in SI units). With this notation, (11) becomes KR =
1 2 Iω . 2
(13)
Although we commonly refer to the quantity 12 Iω 2 as rotational kinetic energy, it is not a new form of energy. It is ordinary kinetic energy because it is derived from a sum over individual kinetic energies of the particles contained in the rigid object. However, the mathematical form of the kinetic energy given by (13) is a convenient one when we are dealing with rotational motion, provided we know how to calculate I. It is important that you recognize the analogy between kinetic energy associated with linear motion 12 mv 2 and rotational kinetic energy 12 Iω 2 . The quantities I and ω in rotational motion are analogous to m and v in linear motion, respectively. (In fact, I takes the place of m every time we compare a linear–motion equation with its rotational counterpart.) The moment of inertia is a measure of the resistance of an object to changes in its rotational motion, just as mass is a measure of the tendency of an object to resist changes in its linear motion. Note, however, that mass is an intrinsic property of an object, whereas moment of inertia depends on the physical arrangement of that mass about a certain axis. Can you think of a situation in which an object’s moment of inertia changes even though its mass does not? Example 2 (The Oxygen Molecule). Consider an oxygen molecule (O2 ) rotating in the xy plane about the z axis. The axis passes through the center of the molecule, perpendicular to its length. The mass of each oxygen atom is m = 2.66 × 10−26 kg, and at room temperature the average separation between the two atoms is d = 1.21 × 10−10 m (the atoms are treated as point masses). (a) Calculate the moment of inertia of the molecule about the z axis. (b) If the angular speed of the molecule about the z axis is Anant Kumar
Mob. No. 8967881837, 9932347531
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Calculation of Moment of Inertia
ω = 4.60 × 1012 rad/s, what is its rotational kinetic energy? Solution: (a) This is a straightforward application of the definition of I. Because each atom is a distance d/2 from the z axis, the moment of inertia about the axis is I =
∑
mi ri2 = m
i
( )2 ( )2 d d 1 +m = md2 2 2 2
1 = (2.66 × 10−26 kg)(1.21 × 10−10 m)2 2 = 1.95 × 10−46 kg·m2 (b) We apply the result we just calculated for the moment of inertia in the formula for KR : 1 2 Iω 2 1 = (1.95 × 10−46 kg·m2 )(4.60 × 1012 rad/s)2 2 = 2.06 × 10−21 J
KR =
4
Calculation of Moment of Inertia
We can evaluate the moment of inertia of an extended rigid object by imagining the object ∑ divided into many small volume elements, each having a mass ∆mi . We use the definition3 I = i ri2 ∆mi and take the limit of this sum as ∆mi → 0. In this limit, the sum becomes an integral over the whole object: ∫ ∑ 2 I = lim ri ∆mi = r2 dm. (14) ∆mi →0
i
It is usually easier to calculate moments of inertia in terms of the volume of the elements rather than their mass, and we can easily make that change by using ρ = m/V , where ρ is the density of the object and V is its volume. We want this expression in its differential form ρ = dm/dV because the volumes we are dealing with are very small. Solving for dm = ρ dV and substituting the result in (14) gives ∫ I=
ρr2 dV.
(15)
If the object is homogeneous, then ρ is constant and the integral can be evaluated for a known geometry. If ρ is not constant, then its variation with position must be known to complete the integration. The density given by ρ = m/V sometimes is referred to as volume density for the obvious reason that it relates to volume. Often we use other ways of expressing density. For instance, when dealing with a sheet of uniform thickness t, we can define a surface density σ = ρt, which signifies mass per unit area. Finally, when mass is distributed along a uniform rod of cross– sectional area A, we sometimes use linear density λ = M/L = ρA, which is the mass per unit length. Example 3. Find the moment of inertia of (a) a thin uniform hoop (Figure 4), and (b) a thin uniform disk (Figure 5) of mass M and radius R about an axis passing through the center and perpendicular to the plane. Solution: (a) All mass elements dm are at the same distance r = R from the axis, and so the moment of inertia ∫ ∫ 2 2 dm = M R2 . I = r dm = R Note that this moment of inertia is the same as that of a single particle of mass M located a distance R from the axis of rotation. 3
Now onwards, r will be used in place of ρ for notational reasons. The symbol ρ will be used to denote the mass density. Anant Kumar
Mob. No. 8967881837, 9932347531
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Calculation of Moment of Inertia y
y
dr
dm
r
dm O
O
x
x
R R
Figure 4: The mass elements dm of a uniform
Figure 5: Calculating the moment of inertia of a
hoop are all the same distance from O.
thin uniform disk.
(b) Take the mass element as a thin ring of radius r < R having width dr as shown in Figure 5. Since the disk is uniform, the total mass M is distributed uniformly over the entire area with a constant surface M . The area of the elementary ring is 2πrdr and its mass is dm = σ2πrdr. Thus the mass density σ = πR2 moment of inertia of this ring about the given axis is (using the result of the part (a)) dI = r2 dm = σ2πr3 dr. Since, moment of inertia is additive, the moment of inertia of the disk about the given axis is found by adding the moment of inertia of the elementary rings whose radius vary from 0 to R. The process is that of integration: ∫ r=R ∫ R M R4 1 R4 I= dI = 2πσ = 2π = M R2 . r3 dr = 2πσ 2 4 πR 4 2 r=0 0
The moments of inertia of rigid bodies with simple geometry (high symmetry) are relatively easy to calculate provided the rotation axis coincides with an axis of symmetry. The calculation of moments of inertia about an arbitrary axis can be cumbersome, however, even for a highly symmetric object. Fortunately, use of an important theorem, called the parallel–axis theorem, often simplifies the calculation. We state it without any proof. ⊙ Theorem 2 (Parallel–axis theorem). Suppose the moment of inertia of a rigid body of a mass M about an axis CC ′ through its center of mass is ICM . Then, the moment of inertia of this body about any axis parallel to CC ′ and a distance d away from it is I = ICM + M d2 .
(16)
Figure 6 illustrates the theorem. One more theorem, which simplifies calculation of moment of inertia of laminar objects is the perpendicular axis theorem. Again, it is stated without any proof with Figure 7 illustrating the idea. ⊙ Theorem 3 (Perpendicular axis theorem). The moment of inertia of a flat, two dimensional rigid body about an axis OZ perpendicular to the plane of the body (O being an arbitrary point in this plane) is the sum of the moments of inertia about any two mutually perpendicular axes OX and OY , both lying in the plane of the body. Example 4. Find the moment of inertia of a homogenous solid sphere of mass M and radius R about a tangent. Solution: The moment of inertia of the sphere about an axis through its center of mass (i.e. diameter) is ICM = 25 M R2 . Applying the parallel axis theorem, the required moment of inertia becomes I = ICM + M R2 = Anant Kumar
7 2 M R2 + M R2 = M R2 . 5 5 Mob. No. 8967881837, 9932347531
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Rotational Dynamics C Z Z
IZ = IX + IY
CM d
Y
M O C′
X
Z′
Figure 6: Illustrating the parallel axis theorem.
Figure 7: The perpendicular axis theorem for a planar object.
5
Rotational Dynamics
5.1
Torque
Why are a door’s doorknob and hinges placed near opposite edges of the door? This question actually has an answer based on common sense ideas. The harder we push against the door and the farther we are from the hinges, the more likely we are to open or close the door. When a force is exerted on a rigid object pivoted about an axis, the object tends to rotate about that axis. The tendency of a force to rotate an object about some axis is measured by a vector quantity → called torque − τ (tau). − → Definition 2 (Torque). Let a force F act at a point P whose position vector measured from − → → a point O be − r (see Figure 8), then the torque of F about the point O is defined as the vector − → → cross product of − r and F : − → − → → τ =− r × F. (17) From this definition, we note the following facts: • Torque is defined about a certain point. → → • The magnitude of the torque, |− τ | = rF sin θ, where θ is the angle between the vectors − r − → and F . Plane containing vectors − → − → r and F
→ − τ
− → F
O
− → r
θ P
Figure 8: The direction of the vector ⃗τ is perpendicular to the plane containing the vectors ⃗r and F⃗ , and is determined by the right–hand thumb rule.
Anant Kumar
Mob. No. 8967881837, 9932347531
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Rotational Dynamics
→ • The direction of the torque is perpendicular to the plane which contains the vectors − r and − → F (that is, the torque is perpendicular to both the position vector and the force) and is determined by the right–hand thumb rule. The magnitude of the torque produced by a force can be interpreted in two manners. Consider the wrench pivoted on the axis through O in Figure 9. With respect to the point O, the magnitude → of the position vector is just r. The magnitude of the torque is |− τ | = rF sin θ. Written in this manner, we see that torque is the product of the magnitude of the position vector and the component F sin θ of the force, that is perpendicular to the position vector. Alternatively, if we note that d = r sin θ is the perpendicular distance of the axis through O from the line of action − → of the force F (The line of action of a force is an imaginary line extending out both ends of the − → vector representing the force. The dashed line extending from the tail of F in Figure 9 is part − → → of the line of action of F .), we can write |− τ | = F d. This quantity d is called the moment arm − → (or lever arm) of F . In case, more than one force acts on a rigid body, the net torque about any point is the vector sum of the individual torques about the same point. Torque should not be confused with force. Forces can cause a change in linear motion, as described by Newton’s second law. Forces can also cause a change in rotational motion, but the effectiveness of the forces in causing this change depends on both the forces and the moment arms of the forces, in the combination that we call torque. Torque has units of force times length — newton · meter in SI units – and should be reported in these units. Do not confuse torque and work, which have the same units but are very different concepts. y
⃗ F
F sin θ
R1
r R
θ
O
θ d
O
F cos θ
Line of action
F⃗1
2
z ⃗2 F
x
Figure 9: It is the component F sin θ that tends to rotate the wrench about O.
Figure 10: A solid cylinder pivoted about the z axis ⃗1 is R1 , and the mothrough O. The moment arm of F ⃗2 is R2 . ment arm of F
Example 5. A one–piece cylinder is shaped as shown in Figure 10, with a core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the drawing (which − → coincides with z axis). A rope wrapped around the drum, which has radius R1 , exerts a force F 1 to the − → right on the cylinder. A rope wrapped around the core, which has radius R2 , exerts a force F 2 downward on the cylinder. (a) What is the net torque acting on the cylinder about the rotation axis? (b) Suppose F1 = 5.0 N, R1 = 1.0 m, F2 = 15.0 N, and R2 = 0.50 m. What is the net torque about the rotation axis, and which way does the cylinder rotate from rest? − → Solution: (a) The torque due to F 1 is −R1 F1 (the sign is negative because the torque tends to produce − → clockwise rotation). The torque due to F 2 is +R2 F2 (the sign is positive because the torque tends to produce counterclockwise rotation). Therefore, the net torque about the rotation axis is ∑ τ = τ1 + τ2 = −R1 F1 + R2 F2 , Anant Kumar
Mob. No. 8967881837, 9932347531
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Rotational Dynamics
the direction being along the positive z axis if the above quantity is positive, along the −z axis otherwise. We can make a quick check by noting that if the two forces are of equal magnitude, the net torque is negative because R1 > R2 . Starting from rest with both forces acting on it, the cylinder would rotate − → − → clockwise because F 1 would be more effective at turning it than would F 2 . (b) ∑ τ = −(5.0 N)(1.0 m) + (15.0 N)(0.50 m) = 2.5 N · m Because the net torque is positive, if the cylinder starts from rest, it will commence rotating counterclockwise with increasing angular velocity. (If the cylinders initial rotation is clockwise, it will slow to a stop and then rotate counterclockwise with increasing angular speed.)
5.2
Relation between Torque and Angular acceleration
In this section we show that the angular acceleration of a rigid object rotating about a fixed axis is proportional to the net torque acting about that axis. Before discussing the more complex case of rigid–body rotation, however, it is instructive first to discuss the case of a particle rotating about some fixed point under the influence of an external force. Consider a particle of mass m rotating in a circle of radius r (Figure 11) under the influence − → − → of a force F that has both a normal component F n along the radius (this normal component must be present to provide the particle the necessary centripetal acceleration), and a tangential − → → component F t . The tangential component provides a tangential acceleration − a t , and Ft = mat . ⃗t F
y
F⃗
d Ft
m
dm O
⃗n F
r r
O
x
Figure 11: A particle rotating in a circle under
Figure 12: A rigid object rotating about an axis
⃗t . A force F ⃗n in the influence of a tangential force F the radial direction also must be present to maintain the circular motion.
through O. Each mass element dm rotates about O with the same angular acceleration α ⃗ , and the net torque on the object is proportional to α ⃗.
− → The torque about the center of the circle due to F t is τ = Ft r = (mat )r. Because the tangential acceleration is related to the angular acceleration through the relationship at = rα, the torque can be expressed as τ = (mrα)r = (mr2 )α. But mr2 is the moment of inertia of the particle about the axis that is perpendicular to the plane of the circle and passes through the center. Thus (taking the directions into account), τ = Iα.
(18)
That is, the torque acting on the particle is proportional to its angular acceleration, and the proportionality constant is the moment of inertia. It is important to note that τ = Iα is the rotational analogue of Newton’s second law of motion, F = ma. Anant Kumar
Mob. No. 8967881837, 9932347531
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Rotational Dynamics
Now let us extend this discussion to a rigid object of arbitrary shape rotating about a fixed axis, as shown in Figure 12. The object can be regarded as an infinite number of mass elements dm of infinitesimal size. If we impose a cartesian coordinate system on the object, then each mass → element rotates in a circle about the origin, and each has a tangential acceleration − a t produced − → by an external tangential force d F t . For any given element, we know from Newton’s second law that dFt = (dm)at . − → → The torque d− τ associated with the force d F t acts about the origin and is given by dτ = rdFt = (rdm)at . Noting that at = rα, the above expression becomes, dτ = (rdm)rα = (r2 dm)α. It is important to recognize that although each mass element of the rigid object may have a → → different linear acceleration − a t , they all have the same angular acceleration − α . With this in mind, we can integrate the above expression to obtain the net torque about O due to the external forces: (∫ ) ∫ ∑ 2 2 τ = (r dm)α = r dm α, where α has been taken out of the integral because it is the same for all mass elements. Since the quantity in the parentheses is the moment of inertia of the rigid body about the axis through O, we obtain (after taking the directions into account): ∑ − → → τ = I− α. (19) So, again we see that the net torque about the rotation axis is proportional to the angular acceleration of the object, with the proportionality factor being I, a quantity that depends upon the axis of rotation and upon the size and shape of the object. Finally, note that the above result also applies when the forces acting on the mass elements have radial components as well as tangential components. This is because the line of action of all radial components must pass through the axis of rotation, and hence all radial components produce zero torque about that axis. Example 6 (Rotating Rod). A uniform rod of length L and mass M is attached at one end to a frictionless pivot and is free to rotate about the pivot in the vertical plane, as shown in Figure 13. The rod is released from rest in the horizontal position. What is the initial angular acceleration of the rod and the initial linear acceleration of its right end? Solution: The only force contributing to torque about an axis through the pivot is the gravitational force → M− g exerted on the rod. (The force exerted by the pivot on the rod has zero torque about the pivot because its moment arm is zero.) To compute the torque on the rod, we can assume that the gravitational force acts at the center of mass of the rod, as shown in Figure 13. The torque due to this force about an axis through the pivot is ( ) L . τ = Mg 2 ∑ With τ = Iα, and I = 31 M L2 , we obtain ( ) ( ) 1 L 3g M L2 α = M g ⇒α= . 3 2 2L All points on the rod have this angular acceleration. To find the linear acceleration of the right end of the rod, we use at = rα with r = L to obtain as the acceleration of the right end as 3 at = g. 2 Anant Kumar
Mob. No. 8967881837, 9932347531
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Rotational Dynamics M
O R
T
T
L/2
Pivot
m
Mg
mg
Figure 13: The uniform rod is piv-
Figure 14: A falling smokestack.
Figure 15: Rotating pulley.
oted at the left end.
This result – that at > g for the free end of the rod – is rather interesting. It means that if we place a coin at the tip of the rod, hold the rod in the horizontal position, and then release the rod, the tip of the rod falls faster than the coin does! Example 7 (Falling Smokestacks). When a tall smokestack falls over, it often breaks somewhere along its length before it hits the ground, as shown in Figure 14. Why does this happen? Solution: As the smokestack rotates around its base, each higher portion of the smokestack falls with an increasing tangential acceleration. (The tangential acceleration of a given point on the smokestack is proportional to the distance of that portion from the base.) As the acceleration increases, higher portions of the smokestack experience an acceleration greater than that which could result from gravity alone; this is similar to the situation described in the last example. This can happen only if these portions are being pulled downward by a force in addition to the gravitational force. The force that causes this to occur is the shear force from lower portions of the smokestack. Eventually the shear force that provides this acceleration is greater than the smokestack can withstand, and the smokestack breaks. Example 8. A wheel of radius R, mass M , and moment of inertia I is mounted on a frictionless, horizontal axle, as shown in Figure 15. A light cord wrapped around the wheel supports an object of mass m. Calculate the angular acceleration of the wheel, the linear acceleration of the object, and the tension in the cord. Solution: The torque acting on the wheel about its axis of rotation is τ = T R, where T is the force exerted by the cord on the rim of the wheel. (The gravitational force exerted by the Earth on the wheel and the normal force exerted ∑ by the axle on the wheel both pass through the axis of rotation and thus produce no torque.) Because τ = Iα, we obtain ∑
τ = Iα = T R,
⇒
α=
TR . I
(1)
Applying Newton’s second law to the block, taking downward direction as positive, we obtain ∑ Fy = mg − T = ma ⇒ T = mg − ma. (2) Substituting this expression for T , expression (1) above, we obtain α=
mgR − maR . I
Because the object and wheel are connected by a string that does not slip, the linear acceleration of the suspended object is equal to the linear acceleration of a point on the rim of the wheel. Therefore, Anant Kumar
Mob. No. 8967881837, 9932347531
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Work, Power and Energy
the angular acceleration of the wheel and this linear acceleration are related by a = Rα. Putting this expression in the last expression and solving for α, we obtain Iα = mgR − mR2 α
⇒
α=
g 1 ). ·( I R 1+ mR2
Therefore, the linear acceleration of the block is a = Rα = ( 1+ obtain the tension as
6 6.1
g I mR2
) . And from expression (2), we
mg ). T =( mR2 1+ I
Work, Power and Energy Work and power
In this section, we consider the relationship between the torque acting on a rigid object and its resulting rotational motion in order to generate expressions for the power and a rotational analog − → to the work-kinetic energy theorem. Refer to Figure 3 again. Suppose that a force F is applied − → → to the point A. Then, for an infinitesimal displacement d− r , the work done by the force F is − → → dW = F · d− r − → − − → → − → → → → = F · ( v dt) = F · (− ω ×→ r )dt (since − v =− ω ×− r) − → − → − → = ( ω × r ) · F dt (since dot product is commutative) − → → − − → → − → → → → =− ω dt · (− r × F) (∵ (− a × b)·− c =→ a ·(b ×− c )) − → − → − → − → − → − → = ( ω dt) · τ = τ · d φ (∵ ω dt = d φ ) − → Thus, the total amount of work done by the torque of the force F is ∫ W = τz dφ,
(20)
where the integration limit is over the entire angular displacement and the z axis is taken along the direction of rotation with the positive direction of coordinate φ coinciding with it. − → The power delivered to the rigid body by the force F is P= Thus, the power delivered is
→ d− φ dW → → → =− τ · =− τ ·− ω. dt dt → → P=− τ ·− ω.
(21)
− → → This expression is analogous to P = F · − v in the case of linear motion.
6.2
Work and energy in rotational motion
In studying linear motion, we found the energy concept — and, in particular, the work-kinetic energy theorem — extremely useful in describing the motion of a system. The energy concept can be equally useful in describing rotational motion. From what we learned of linear motion, we expect that when a symmetric object rotates about a fixed axis, the work done by external forces equals the change in the rotational energy.
Anant Kumar
Mob. No. 8967881837, 9932347531
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Work, Power and Energy
To show this, consider a rigid body rotating about the z axis. We have ∑
dωz dωz dφ dωz =I =I ωz . dt dφ dt dφ ∑ Rearranging this expression and noting that τ dφ = dW , we obtain ∑ τ dφ = dW = Iωz dωz . τ = Iα = I
Integrating this expression, we get for the total work done by the net external force acting on a rotating system ∫ ∫ ωzf 1 2 1 2 W = dW = − Iωzi , (22) Iωz dωz = Iωzf 2 2 ωzi where the angular velocity component about the z axis changes from ωzi and ωzf over the total displacement. Thus, the net work done by external forces in rotating a symmetric rigid object about a fixed axis equals the change in the object’s rotational energy.
E i = U = MgL/2
O
R
L/2 O′
m2 h h m1
1 E f = K R = – Iω ω2 2
Figure 16: A uniform rigid rod pivoted at O rotates in a vertical plane under the action of gravity.
Figure 17: Connected pulleys.
Example 9. A uniform rod of length L and mass M is free to rotate on a frictionless pin passing through one end (Fig 16). The rod is released from rest in the horizontal position. (a) What is its angular speed when it reaches its lowest position? (b) Determine the linear speed of the center of mass and the linear speed of the lowest point on the rod when it is in the vertical position. Solution: (a) When the rod is horizontal, it has no rotational energy. The potential energy relative to the lowest position of the center of mass of the rod (O′ ) is M gL/2. When the rod reaches its lowest position, the energy is entirely rotational energy, 21 Iω 2 , where I is the moment of inertia about the pivot. Because I = 31 M L2 , and the mechanical energy is conserved (there are no dissipative forces), we have Ei = Ef , or √ ( ) 1 2 1 1 1 3g 2 2 M gL = Iω = ML ω ⇒ ω= . 2 2 2 3 L (b) Using the relation v = rω between linear and angular velocity, we obtain vCM =
L 1√ 3gL , ω= 2 2
and for the lowest point v = Lω =
√
3gL .
Example 10. Consider two cylinders having masses m1 and m2 , where m1 ̸= m2 , connected by a string passing over a pulley, as shown in Figure 17. The pulley has a radius R and moment of inertia I about its axis of rotation. The string does not slip on the pulley, and the system is released from rest. Find the linear speeds of the cylinders after cylinder 2 descends through a distance h, and the angular speed of the Anant Kumar
Mob. No. 8967881837, 9932347531
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Work, Power and Energy
pulley at this time. Solution: We are now able to account for the effect of a massive pulley. Because the string does not slip, the pulley rotates. We neglect friction in the axle about which the pulley rotates for the following reason: Because the axle’s radius is small relative to that of the pulley, the frictional torque is much smaller than the torque applied by the two cylinders, provided that their masses are quite different. Mechanical energy is constant; hence, the increase in the system’s kinetic energy (the system being the two cylinders, the pulley, and the Earth) equals the decrease in its potential energy. Because Ki = 0 (the system is initially at rest), we have ∆K = Kf − Ki = ( 12 m1 vf2 + 21 m2 vf2 + 12 Iωf2 ) − 0 where vf is the same for both blocks. Because vf = Rωf , this expression becomes ( ) 1 I ∆K = m1 + m2 + 2 vf2 . 2 R From Figure 17, we see that the system loses potential energy as cylinder 2 descends and gains potential energy as cylinder 1 rises. That is ∆U2 = −m2 gh, and ∆U1 = m1 gh. Applying the principle of conservation of energy in the form ∆K + ∆U1 + ∆U2 = 0, gives ( ) I 1 m1 + m2 + 2 vf2 + m1 gh − m2 gh = 0 2 R [ ] 12 2(m2 − m1 )gh ) ⇒ vf = ( m1 + m2 + RI2 Because vf = Rωf , the angular speed of the pulley at this instant is [ ] 12 vf 1 2(m2 − m1 )gh ( ) . ωf = = R R m1 + m2 + RI2
Anant Kumar
Mob. No. 8967881837, 9932347531
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Work, Power and Energy
R
Hoop or cylindrical shell I CM =MR 2
Solid cylinder or disk =1 MR 2 CM 2
R
Hollowcylinder 1 I CM = M(R 12 +R 22) 2
R1
R2
Rectangular plate I CM =1 M(a2 +b2) 12 b a
Long thin rod with rotation axis through center I CM = 1 ML 2 12
Long thin rod with rotation axis through end
L
I = 1 ML 2 3
Solid sphere I CM =2 MR 2 5
Thin spherical shell I CM =2 MR 2 3 R
L
R
Figure 18: The moments of inertia of some common objects about the standard axes of symmetry.
Anant Kumar
Mob. No. 8967881837, 9932347531