Physics of Rotation Physics 109, Introduction To Physics Fall 017
Outline Next two lab periods Rolling without slipping Angular Momentum Comparison with Translation New Rotational Terms Rotational and Tangential Motion Kinetic Energy and Rotational Inertia Rolling Experiment
Rolling Without Slipping We will compare the rolling speed of several different objects going down an inclined plane Note that there is a description in the 11 Lab Manual-- We are going to perform the experiment in a different way We will not use the computer and sensor to monitor the motion down the plane (too much opportunity for error) We will use a meter stick and a stop watch to measure the rolling objects.
Angular Momentum This is one of the best ways to demonstrate the principle of conservation of angular momentum Every participant will experience this phenomenon It is here that we will have to use the vector cross product and the right-hand rule We will cover these in the lecture prior to the lab period. Prepare to be amazed!
Translation and Rotation We have been dealing with changes in position which we call Translations There are also possibilities for angular changes and this is called Rotational Motion The rolling object experiment involves both as the objects roll down an inclined plane Please note that we are skipping Chapter 7 in the text which deals with vibrations we will return to this topic later.
Pure Rotation We must become used to expressing angles in terms of RADIANS. The angle in Radians, θ, is expressed as the arc length divided by the radius. θ r l θ = l/r Note that radians are dimensionless!!
New Rotational Terms Angular displacement, θ = θ Angular velocity = ω = θ/ t units radians/sec. θ0 units-radians At constant angular velocity each rotation takes the same amount of time... this is the Period, T of the rotation. So for constant ω Angular acceleration = α = ω/ t = radians/sec = π/t
Kinematic Equations These are for rotational motion at constant acceleration... Translational Rotational Vx = vx0 + axt ω = ω0 + αt X = x0 + v0t + 1/at θ = θ0 + ω0t + 1/αt Vx = vx0 + ax x ω = ω0 + α θ
Rotational Analogs Translational Rotational Position, x Angular Position, θ Displacement, x Displacement, θ Velocity, vx Angular velocity, ω Acceleration, a Time, t Angular acceleration, α Time, t
Rotational and Tangential Motion Tangential velocity and speed, given T the period of rotation Vt = distance/time = πr/t = πr/π/ω = rω This is called the tangential speed. For the CD, outer radius, rω = 1.31 m/s and for the inner radius, rω = 1.5 m/s Centripetal Acceleration = vt /r = (rω) /r = rω This is valid for radian measure
Kinetic Energy and Rotational Inertia The kinetic energy, K, of a small mass at a radius r, with a tangential velocity of vt = ½ mvt For a collection of such point masses in rotation: 1 1 K = [ ] mi v ti = [ ]mi r i ω So we can write: K = 1 ( m r )ω i i And we call the sum the Rotational Inertia The symbol is I, so K = ½ Iω
Rotational Inertia Be sure to look at the information given on page 179 of your text. The I values for a number of different shapes are given there. For purposes of our experiment we will need four different values... Hollow cylinder about its axis... I = MR A disk or solid cylinder about its axis, I = ½ MR A hollow sphere about its diameter I = /3 MR A solid sphere about its diameter I = /5 MR
Rotational Inertias
Shapes we will use (+1)
The following factors apply to the ball rolling down the plane: P A. The angle of the plane B. The force of gravity C. Friction at point P D. A and B E. A, B and C
Point P has something in common with: P A. A block sliding down the plane B. A ball thrown into the air C. skier going down the slope D. None of the above
The common factor is: P A. Rolling motion B. An instant of zero velocity. C. normal force D. None of the above
A CD maintains a constant linear rate as it rotates. How does it accomplish this? A. The head moves out faster at the rim. B. It changes RPM as the distance from the center changes. C. Neither of the above. D. Both of the above.
A shape you will see again The formula given here is for the rotational Inertia. There is another related property called the Moment of Inertia, or first moment. It does not contain the mass term, but does contain the area. The formula becomes : I = 1/1 ba3
And where will you see it? The bending of a beam involves knowledge of the bending moment M and the moment of inertia of the beam cross section, I. The stress in the outer portion of the beam is given by σ = Mc/EI where c is the beam half height and E is the elastic modulus of the material. L/ P c h P/ M= P(L/), I = 1/1bh P/ 3 b
An Aside: Why do you think that a popular beam cross section is called an I-Beam? Because most of the material is far from the center line, where it is most resistant to bending.
Another consideration: We need to use the Parallel Axis Theorem the object is rotating about the point that is in contact with the surface, a rotation axis that is parallel to the object axis, so the I for this situation is I = mr + Icm = mr (1 + k) where k is a number between 0 and 1, and depends on the nature of the object.
Kinematics of Rolling A rolling object combines translational and rotational motion We can break the motion into the translation of the center of mass and rotation about the center of mass One full turn of a disk moves the disk a distance of πr. The angular displacement θ = π. So, the vcm = πr/ t, where t is the time for a full rotation.
Kinematics of Rolling (cont) Now the disk angular speed is ω = π/ t so: Combine this with the previous expression vcm = πr/ t And obtain vcm = ωr So, for an object that rolls without slipping, there is a direct relationship between the translational speed and the rotation rate.
Kinetic Energy in Rolling The total kinetic energy is the sum of the translational kinetic energy and the rotational kinetic energy. This is shown below: 1 1 K rolling = m v cm+ I cm ω We will repeat the experiment performed by Galileo, first lets calculate the kinetic energy of a rolling solid ball: v cm 1 1 K rolling = mv cm m R 5 R
KE in rolling (cont) This simplifies to: 1 1 7 K rolling = m v cm + m v cm = m v cm 5 10 As the ball rolls down the ramp, its potential energy, mgh is translated into kinetic energy. 7 10gh mgh= mv cm so v cm= ( ) 10 7 Note that the result is independent of the mass and the radius of the ball!
The Race of the Rolling Bodies Various round, rigid bodies are started rolling down a plane together. Which will reach the bottom first? Using conservation of energy: K1 U 1 K U 0 Mgh 1 Mv Mgh 1 Mv cm cm 1 I 1 cmr v cm R
The Race of the Rolling Bodies () From the previous chart: Mgh 1 1 c Mv cm So the speed at the bottom of the incline is: v cm gh 1 c 0.5 Note that the radius and mass of the body drops out, and the remaining factor is the multiplier for moment of inertia!
The Race of the Rolling Bodies (3) So, comparing the multiplier for moment of inertia, we have: Solid sphere, c = /5 Solid cylinder, c = ½ Hollow sphere, c = /3 Hollow cylinder, c = 1 As a result, this will be the order of finish.
The Experiment: We will use a meter stick and a stop watch, and measure the time each object rolls down the ramp for one meter. We will not assume the k multiplier of the rolling inertia I for each object, but will calculate it based on the rolling time. We will take multiple time measurements for each object, and use the average to calculate k.
The calculation: v cm 1 1 K rolling = mv cm+ ( m R )( ) 5 R Re write this equation: v cm 1 1 K rolling = mv cm k m R R gh=v cm 1 k gh v = =(( x 1ength(m))/t ) (1+ k ) cm Finally, on the next slide:
The relation between t and k: t= length(m)/ (gh /(1+ k )) Where h is the height difference between the top of the ramp and the bottom And t is the time you will measure for the object to roll. You will determine k for each object, and compare it with the theoretical value. (and of course, calculate the percent error)
The Race of the Rolling Bodies Various round, rigid bodies are started rolling down a plane together. Which will reach the bottom first? Using conservation of energy: K1 U 1 K U 0 Mgh 1 Mv Mgh 1 Mv cm cm 1 I 1 cmr v cm R
The Race of the Rolling Bodies () From the previous chart: Mgh 1 1 c Mv So the speed at the bottom of the incline is: v cm cm gh 1 c 0.5 Note that the radius and mass of the body drops out, and the remaining factor is the multiplier for moment of inertia!
What will be the order of finish? A. solid cylinder, hollow cylinder, hollow sphere, solid sphere. B. hollow cylinder, solid sphere, solid cylinder, hollow sphere. C. solid sphere, solid cylinder, hollow sphere, hollow cylinder. D. hollow sphere, solid sphere, hollow cylinder, solid cylinder.
We calculate the moment of inertia about point P using: P A. The moment of inertia about the axis of the cylinder. B. The Parallel Axis Theorem C. Both of the above. D. Neither of the above
One of the cylinders has a higher moment of inertia about its axis? The masses and diameters are the same. Why? A. More of the mass is concentrated at the center. B. More of the mass is concentrated at the rim.. C. Not enough information to tell.
How do we calculate angular speed? A. Divide the angular change by the time B.. Divide the number of revolutions by the time C. Both D. Neither