Rotational Motion Notes

Rotational Motion

Units of Chapters 8 & 9

  • Angular Quantities
  • Constant Angular Acceleration
  • Rolling Motion (Without Slipping)
  • Torque
  • Statics: The Conditions for Equilibrium and Solving Statics Problems
  • Rotational Dynamics; Torque and Rotational Inertia
  • Solving Problems in Rotational Dynamics
  • Rotational Kinetic Energy
  • Angular Momentum and Its Conservation
  • Vector Nature of Angular Quantities

Concept Test 8.1a: Bonnie and Clyde I

  • Scenario: Bonnie sits on the outer rim of a merry-go-round, and Clyde sits midway between the center and the rim. The merry-go-round makes one complete revolution every two seconds.
  • Question: Clyde’s angular velocity is:
  • Answer: C) same as Bonnie’s

Angular Quantities

  • In purely rotational motion, all points on the object move in circles around the axis of rotation (“O”). The radius of the circle is r.
  • All points on a straight line drawn through the axis move through the same angle in the same time.

Angular Quantities - Angle in Radians

  • The angle \\theta in radians is defined as: \theta = \frac{s}{r} , where s is the arc length. (8-1a)
  • Units of \\theta: Radians
  • Conversion: 2\\pi \text{ rad} = 360^0

Angular Quantities - Angular Displacement

  • Linear displacement: \\Delta x = x2 – x1
  • Angular displacement: \\Delta \theta = \theta2 - \theta1
  • Connection: x = r\\theta, where x is the arc length.

Angular Quantities - Average Angular Velocity

  • Average linear velocity: v = \frac{\\Delta x}{\\Delta t}
  • Average angular velocity (\omega: omega) is defined as the total angular displacement divided by time: \omega = \frac{\\Delta \theta}{\\Delta t} (8-2a)
  • Recall: v = r\\omega, where v is the tangential speed.

Angular Quantities - Tangential Speed

  • Every point on a rotating body has the SAME angular velocity \omega and a linear velocity v given by: v = r\\omega (8-4)
  • Do all points along a radius have the same tangential speed? NO!
  • Objects farther from the axis of rotation will move faster.

Angular Quantities - Right Hand Rule

  • The direction of angular velocity (\omega) is given by a Right Hand Rule (“Hitch Hiker”).
  • If you curl your fingers in the direction of the rotation, your thumb will point in the direction of \omega.
  • Direction notation: “dot” (out of the page) and “x” (into the page).
  • \omega is perpendicular to the plane of rotation.

Vector Nature of Angular Quantities

  • The angular velocity vector points along the axis of rotation; its direction is found using a right hand rule.

Angular Quantities - Instantaneous Angular Velocity

  • The instantaneous angular velocity: \omega = \lim_{\\Delta t \to 0} \frac{\\Delta \theta}{\\Delta t} (8-2b)
  • What happens if \omega changes? What quantity will we have?

Angular Quantities - Angular Acceleration

  • Average linear acceleration: a = \frac{\\Delta v}{\\Delta t}
  • The angular acceleration (\alpha: alpha) is the rate at which the angular velocity changes with time: \alpha = \frac{\\Delta \omega}{\\Delta t} (8-3a)
  • Connection: a = r\\alpha, where a is the tangential acceleration.
    • If \omega is increasing, \alpha points in the direction of \omega.
    • If \omega is decreasing, \alpha points opposite to the direction of \omega.
    • \alpha is perpendicular to the plane of rotation.

Vector Nature of Angular Quantities - Angular Acceleration

  • Angular acceleration points along the axis of rotation.

Angular Quantities - Instantaneous Angular Acceleration

  • The instantaneous angular acceleration: \alpha = \lim_{\\Delta t \to 0} \frac{\\Delta \omega}{\\Delta t} (8-3b)

Linear and Rotational Quantities

LinearTypeRotationalRelation
xdisplacement\thetax = r\\theta
vvelocity\omegav = r\\omega
a_{tan}acceleration\alphaa = r\\alpha

Angular Quantities - Frequency and Period

  • The frequency is the number of complete revolutions per second.
  • Frequencies are measured in hertz.
  • The period is the time one revolution takes: T = \frac{1}{f} (8-8)
  • \omega is also known as Angular Frequency.

Constant Angular Acceleration

  • The equations of motion for constant angular acceleration are the same as those for linear motion, with the substitution of the angular quantities for the linear ones.

Constant Angular Acceleration - Example

  • A wheel 33 cm in diameter accelerates uniformly from 240 rpm to 360 rpm in 6.5 s. How far will a point on the edge of the wheel travel in this time? How far will a point that is halfway from the center travel in this time?

Constant Angular Acceleration - Example 2

  • The tires of a car make 65 revolutions as the car reduces its speed uniformly from 95 km/h to 45 km/h. The tires have a diameter of 0.80 m.
    • (a) What was the angular acceleration of the tires?
    • (b) Assuming a constant \alpha, how long does it take the car to stop?
  • If the car is traveling in the direction indicated, what is the direction of \alpha?

Concept Test 8.3a: Angular Displacement I

  • An object at rest begins to rotate with a constant angular acceleration. If this object rotates through an angle \theta in a time t, through what angle did it rotate in a time \frac{1}{2}t?
  • B) \frac{1}{4} \theta
  • The angular displacement is \theta = \frac{1}{2} \alpha t^2 (starting from rest), and there is a quadratic dependence on time. Therefore, in half the time, the object has rotated through one-quarter of the angle.

Rolling Motion (Without Slipping)

  • Rolling is rotation with translation.
  • Given pure rotation, with angular speed \omega, the speed of point “p” on the rim is v_p \text{ about cm} = r\\omega

Rolling Motion (Without Slipping) - Center of Mass

  • Rolling also involves translation of the CM.
  • Find an expression for how v_{cm} relates to the rotation in terms of \omega.
  • Consider the point of contact between the ground and the rolling wheel (k) instantaneously at rest.
  • Determine how the CM moves about this point: v_{cm}\text{ about k} = r\\omega

Rolling Motion (Without Slipping) - Top of Rim Speed

  • v_{cm} = r\\omega
  • The speed of a point on the top of the rim (about point k) is given by: v{top \text{ about k}} = (2r)\omega = 2(r\\omega) = 2v{cm}

Concept Test 8.4: Using a Wrench

  • You are using a wrench to loosen a rusty nut. Which arrangement will be the most effective in loosening the nut?
  • C) is the most effective because it offeres the largest lever arm.
  • Since the forces are all the same, the only difference is the lever arm. Thus, the arrangement with the largest lever arm (case C) will provide the largest torque.

Torque

  • To make an object start rotating, a force is needed; the position and direction of the force matter as well.
  • If this force produces a rotation, a torque has been exerted on the rotating object.
  • Connection: Torque is the rotational analog of Force.

Torque - Lever Arm

  • The perpendicular distance from the axis of rotation to the line along which the force acts is called the lever arm.

Torque - Lever Arm Examples

  • Here, the lever arm for FA is the distance from the knob to the hinge; the lever arm for FD is zero; and the lever arm for F_C is as shown:
  • A longer lever arm is very helpful in rotating objects.

Concept Test 8.7: Cassette Player

  • When a tape is played on a cassette deck, there is a tension in the tape that applies a torque to the supply reel. Assuming the tension remains constant during playback, how does this applied torque vary as the supply reel becomes empty?
  • C) torque decreases
  • As the supply reel empties, the lever arm decreases because the radius of the reel (with the tape on it) is decreasing. Thus, as the playback continues, the applied torque diminishes in magnitude.

Torque - Definition

  • The torque is defined as: \tau = rF sin \theta (8-10a)
    • Pick \sum \tau about a strategic point.
    • Determine the angle \theta between r and F.
    • A Right Hand Rule gives the direction of \tau.
    • \sin \theta yields the \perp nature of the two vectors.
  • r is the distance from AoR to the location of the force.
  • r_\\perp = r \sin \theta is the \perp distance from AoR to the force’s line of action.

Torque - Summation

  • Just as we sum forces, we also sum Torques.
  • Torques in the direction of \alpha are positive and opposite to \alpha are negative. If there is no \alpha, then the choice is up to you.
  • Point “o” is picked to have the largest number of forces acting on it (so long as you don’t need to find one of them).

Conditions for Equilibrium

  • An object with forces acting on it, but that is not moving, is said to be in equilibrium.

Conditions for Equilibrium - First Condition

  • The first condition for equilibrium is that the forces along each coordinate axis add to zero.
  • \sum F_x = 0
  • \sum F_y = 0
  • \sum F_z = 0

Conditions for Equilibrium - Second Condition

  • The second condition for equilibrium is that there be no torque around any axis; the choice of axis is arbitrary.
  • \sum \tau = 0
  • If these two conditions hold, two statements can be made:

Solving Statics Problems

  • Choose one object at a time, and make a free-body diagram showing all the forces on it and where they act.
  • Choose a coordinate system and resolve the forces into their components.
  • Write equilibrium equations for the forces. If another equation is needed, sum the torques.
  • Choose any axis perpendicular to the plane of the forces and write the torque equilibrium equation. A clever choice here can simplify the problem enormously.
  • Solve.

Solving Statics Problems - Example 1

  • Three children are trying to balance on a seesaw, which consists of a fulcrum rock, acting as a pivot at the center, and a very light board 3.6 m long.
  • Boy A has a mass of 50 kg, and girl B a mass of 35 kg. Where should girl C, whose mass is 25 kg, place herself so as to balance the seesaw?
  • Solution:
    • m_1 = 50 kg
    • m_2 = 35 kg
    • m_3 = 25 kg
    • L = 3.6 m
    • Place m_3 to balance; x = ?
    • Pick spot “o” strategically
    • Draw FBD: 5 forces

Solving Statics Problems - Example 2

  • A uniform 1500 kg beam, 20.0 m long, supports a 15,000 kg printing press 5.0 m from the right support column. Find the force on/of each of the vertical support columns.
  • Solution:
    • Pick \tau inward as positive
    • Pick pt o at left end
    • \sum \taui = (1500g)(10m)(\sin 90) + (15000g)(15m) \sin(90) - (FR)(20m)(\sin 90) = 0
    • = 147,000 + 2,205,000 – 20F_R = 0
    • F_R = 117,600 N
    • Now get F_A
    • \sum Fy = FA + F_B - 1500g - 15,000g = 0
    • F_A = 14,700 + 147,000 - 117,600 = 44,100 N

Solving Statics Problems - Example 3

  • The system is in static equilibrium. Determine FA and FB if the beam has a mass of 1200 kg.
  • Choose “inward” as the positive torque direction.
  • Since F_A comes out negative, it just means that it’s in the opposite direction from the one we chose.
  • This is trivial to fix, so don’t worry about getting all the signs of the forces right before you start solving.
  • F_A pulls downward (via bolts, glue, etc.) for the beam to be in equilibrium.
  • Now find F_B.

Solving Statics Problems - Example 4

  • A 3 m long uniform beam, with a mass of m = 30 kg, is mounted via a hinge on a wall as seen. The beam is held in a horizontal position by a cable that makes an angle of \theta = 25^0 with the beam. The beam supports a sign with a mass of M = 37 kg that is suspended from its end.
  • Find the components of the hinge force exerted on the beam, as well as the tension in the cable.
  • Solution:
    • \sum \tau_{Hinge} = (1.5m)(30g) \sin 90 + (3m)(37kg)g \sin 90 - (3m)T \sin 25 = 0
    • 441 + 1087.8 = 1.268T T = 1,206.7 N
    • \sum F{x \text{ beam}} = F{Hx} - Tx = 0 F{Hx} = T \cos 25 = 1093.6 N
    • \sum F{y \text{ beam}} = F{Hy} + Ty - mg - Mg = 0 Fy = -509.97 + 294 + 362.6 = 146.6 N

Solving Statics Problems - Example 5

  • A young lady (m = 45 kg) ascends a 12 m ladder (M = 30 kg) positioned against a wall. The ladder makes an angle \theta = 70^0 with the ground as shown in the diagram. The coefficient of static friction between the ladder and the ground is 0.25, and there is no friction between the ladder and the wall.
  • Determine how high up the ladder she can ascend before it begins to slip.
  • Choose “inward” as the positive torque direction.
  • (from the base of the ladder)

Table Torque

  • A rectangular table of uniform mass M and length L is held up by two supports \frac{1}{4}L from either end.
    • a) Find the force needed when applied at one end to tip the table.
    • b) Find the same force needed to tip the table when the supports are \frac{1}{8}L from either end.

Rotational Dynamics; Torque and Rotational Inertia

  • Knowing that \tau = rF, we see that since: \alpha = \frac{\tau}{mr^2} (8-11)
  • \alpha is proportional to the torque applied, over the factor mr^2:
  • Angular acceleration is inversely proportional to a spatial dependence on mass!

Rotational Dynamics; Torque and Rotational Inertia - Extended Object

  • This is for a single point mass; what about for an extended object?
  • Since the angular acceleration is the same for the whole object, we can write: \tau = (\sum mr^2)\alpha (8-12)
  • …as if it’s made of little chunks of mass m.

Rotational Dynamics; Torque and Rotational Inertia - Rotational Inertia

  • Just as Inertia was a property of matter resisting a change in linear motion, Rotational Inertia (a.k.a. Moment of Inertia) is a property of matter that resists a change in rotation, and possesses a spatial dependence on mass.
  • The quantity \sum mr^2 is called the Rotational Inertia (Moment of Inertia) of an object.

Rotational Dynamics; Torque and Rotational Inertia - Mass Distribution

  • The distribution of mass matters here – these two objects have the same mass, but the one on the left has a greater Rotational Inertia, since so much of its mass is far from the axis of rotation.
  • Ex. Wiggling your pen.
  • Note: Rotation is harder when mass is farther from the axis of rotation.

Concept Test 8.9: Moment of Inertia

  • Two spheres have the same radius and equal masses. One is made of solid aluminum, and the other is made from a hollow shell of gold. Which one has the bigger moment of inertia about an axis running through its center?
  • b) hollow gold
  • Moment of inertia depends on mass and its distance from the axis squared. It is bigger for the shell since its mass is located farther away from the center.

Rotational Dynamics; Torque and Rotational Inertia - Parallel Axis Theorem

  • The rotational inertia of an object depends not only on its mass distribution, but also the location of the axis of rotation – compare (f) and (g), for example.
  • Parallel Axis Theorem:
    • you can find the rotational inertia of an object about any parallel axis via: I = I_{CM} + Md^2
    • where:
      • d = distance from the CM axis to a new axis parallel to the CM axis.

Solving Problems in Rotational Dynamics

  • Draw a diagram.
  • Decide what the system comprises.
  • Draw a free-body diagram for each object under consideration, including all of the forces acting on it and where they act.
  • Find the axis of rotation and calculate the torques around it.

Solving Problems in Rotational Dynamics - Newton's Second Law

  • Apply Newton’s second law for rotation. (If the rotational inertia is not provided, you need to find it before proceeding with this step.)
  • Apply Newton’s second law for translation and other laws and principles as needed.
  • Solve.
  • Check your answer for units and correct order of magnitude.

Solving Problems in Rotational Dynamics - Example 1

  • A 4 kg, L = 12 m thin rod rotates about an axis through the center, perpendicular to its length. If F1 = 12 N and acts into the page, and F2 = 3 N and acts into the page, what is the angular acceleration?

Solving Problems in Rotational Dynamics - Example 2

  • A pulley (solid cylinder) of mass M and radius R is pivoted about its CM. A very light string is wrapped around the pulley, and a mass m is dangling as seen. If the system is frictionless, determine both the acceleration of the mass, and the angular acceleration of the pulley.
  • Givens: M, m, and R as seen in the diagram.
  • Now proceed as before to find a and \alpha.

Solving Problems in Rotational Dynamics - Example 3

  • Show that the end of a plank of length L and mass M, starting from an angle of \\Phi (with respect to the table), can fall with an acceleration greater than “g”.

Rotational Kinetic Energy

  • The kinetic energy of a rotating object is given by: KE = \frac{1}{2}I\omega^2
  • By substituting in rotational quantities, we find that the rotational kinetic energy can be written as:
    translational and rotational kinetic energy: KE = \frac{1}{2}Mv^2 + \frac{1}{2}I\omega^2 (8-16)

Rotational Kinetic Energy - Conservation of Energy

  • When using conservation of energy, both rotational and translational kinetic energy must be taken into account.
  • All these objects have the same potential energy at the top (same mass), but the time it takes them to get down the incline depends on how much rotational inertia they have:

Rotational Kinetic Energy - Cylinder Example

  • Compare the final velocity of a solid cylinder sliding down the incline to one rolling down the incline. Both have the same mass M and start from rest.

Rotational Kinetic Energy - Work

  • The torque does work as it moves the wheel through an angle \theta:
    • Linear: W = F\Delta x
    • Rotational: W = \tau \Delta \theta (8-17)
    • \\Delta KE{Rot} = \frac{1}{2}I(wf^2 - w_i^2)

Angular Momentum and Its Conservation

  • In analogy with linear momentum, we can define angular momentum L: L = I\omega (8-18)
  • We can then write the total torque as being the rate of change of angular momentum: \tau = \frac{\Delta L}{\Delta t}
  • If the net torque on an object is zero, the total angular momentum is constant:
    • Li = Lf
    • Ii wi = If wf

Vector Nature of Angular Quantities

  • Angular acceleration and angular momentum vectors also point along the axis of rotation.
  • L is in the direction of \omega, also given by the “Hitch Hiker” right hand rule.

Angular Momentum and Its Conservation - Rotational Inertia

  • Therefore, systems that can change their rotational inertia through internal forces will also change their rate of rotation:
    • Ii wi = If wf

Angular Momentum and Its Conservation - Net Work Example

  • Determine the net work if the mass of the disk is M = 10 kg, the mass of the rod is M = 10 kg, and the radius of the disk is R = 1 m.

  • Solution

    • Considering both the disk & rod as one system, there are no Net External Torques. So Angular Momentum is conserved.
    • w_i = 2.4 \frac{\text{rev}}{s} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 15.08 \frac{\text{rad}}{s}
    • w_f = 1.44 \frac{\text{rev}}{s} \times \frac{2\pi \text{ rad}}{1 \text{ rev}} = 9.05 \frac{\text{rad}}{s}
    • \frac{1}{2} Ii wi^2 - \frac{1}{2} If wf^2
    • W_{netRot} = Change in Rot. KE
    • = [\frac{1}{2}(\frac{1}{2}MR^2)(15.08)^2] - [\frac{1}{2}(\frac{1}{2}MR^2 + \frac{1}{12}M(2R)^2)(9.05)^2]
    • = [\[\frac{1}{4}(10)(1^2)(15.08)\^2\] -\ [\frac{1}{2}\times568.5\ ]

  • Negative indicates M.E. is lost, slows down

Additional Examples

  • Rolling Cylinder
  • Falling Meter Stick

Summary of Chapters 8 & 9

  • Angles are measured in radians; a whole circle is 2\pi radians.
  • Angular velocity is the rate of change of angular position.
  • Angular acceleration is the rate of change of angular velocity.
  • The angular velocity and acceleration can be related to the linear velocity and acceleration.
  • The frequency is the number of full revolutions per second; the period is the inverse of the frequency.

Summary of Chapters 8 & 9 - Continued

  • The equations for rotational motion with constant angular acceleration have the same form as those for linear motion with constant acceleration.
  • Torque is the product of force and lever arm.
  • The rotational inertia depends not only on the mass of an object but also on the way its mass is distributed around the axis of rotation.
  • The angular acceleration is proportional to the torque and inversely proportional to the rotational inertia.

Summary of Chapters 8 & 9 - Continued 2

  • An object that is rotating has rotational kinetic energy. If it is translating as well, the translational kinetic energy must be added to the rotational to find the total kinetic energy.
  • Angular momentum is: L = I\omega
  • If the net torque on an object is zero, its angular momentum does not change.