ROTATIONAL MOTION FROM TRANSLATIONAL MOTION Velocity Acceleration 1-D otion 3-D otion Linear oentu TO
We have shown that, the translational otion of a acroscopic object is equivalent to the translational otion of the CM But this result is not sufficient to describe the otion of a acroscopic object
Case A Case B
The CM is stationary in both cases Fast spinning slow spinning
TORQUE Notice: causes no rotation does cause rotation TOP VIEW No rotation Notice, in this case only the coponent F 2 sinα is responsible for the rotation
LATERAL VIEW TOP VIEW A B F F A B Notice also, a force F acting on point B is ore effective in rotating the disk than the sae force F acting on point A
Hence, to attain a ore effective rotation,
F
The vector product Z X Y = = = + - +
= = = =
Exaple = = X Z Y = =
Practice =
Practice = = = = = =
Let decopose the vector r into two other vectors, one perpendicular to F and another parallel to F r = a + b
DESCRIPTION of ROTATIONAL MOTION METHOD: Applying the conservation of echanical energy CASE: A rotating acroscopic object Rotation Translation Translation A B How do observers A and B apply the concept of conservation of echanical energy?
chapters One specific proble we have in ind is the following: a cylinder rolling down Y h o When applying the conservation of echanical energy, if the rotational otion were neglected, we would state, ½ v 2 CM + gy = g h o 1 But this expression is not correct.
We can use the energy of isolated syste to treat class of probles concerning the rolling otion of a rigid body down a rough incline. In these cases, the gravitational potential energy of the object-earth syste decreases the rotational and translational kinetic energies of the object increase. For exaple, consider the sphere rolling without slipping after being released fro rest at the top of the inclined. Note that the accelerated rolling otion is possible only if a friction force is present between the the sphere and the incline to produce a net torque about the center of ass. Despite the presence of friction, no loss of echanical energy occurs because the contact point is at rest relative to the surface at any instant. (On the other hand, if the sphere were to slip, echanical energy of the sphere-incline-earth syste would be lost due to the kinetic friction force) Ref: Principes of Physics, by Serway and Jewett
Kinetic Energy of a Syste of Particles
Z' Z Y' Y X' X r i "i"
K = Kinetic energy = = = this is justified in the next page K = Kinetic energy easured fro the reference XYZ Velocity of the CM with respect to XYZ Kinetic energy easured fro the reference X'Y'Z' (whose origin is attached to the CM of the particles.)
Justification of i i v i = 0 Notice, the position of the center of ass with respect to XYZ is given by r CM = i i r i / i i O r i r i r CM O O is the CM of the syste of particles r CM = i i r i / i i Since the point O is the center of ass of the particles, then r CM is the vector zero. That is, 0 = i i r i / i i, or 0 = i i r i Taking the derivative with respect to tie 0 = i i v i which is what we wanted to justify
Exaple Y' X' Rolling cylinder Y X K = ½ M v 2 + K CM Kinetic energy of the cylinder evaluated fro the reference XYZ Kinetic energy of the cylinder evaluated fro the reference X'Y'Z' Exaple
Kinetic Energy K CM of a Rotating Rigid Body K CM = Z' ω Z R i CM r i ' v i ' X Y Z' K CM = K =
Exaple: I = I = I = MR 2
EXAMPLE 1: Consider two asses joined by a asless rod. d a) What is the oentu of inertia with respect to an axis that passes through the center of ass and that is perpendicular to the rod? I CM = (d/2) 2 + (d/2) 2 = d 2 / 2 b) What is the oentu of inertia with respect to an axis passing through one end of the rod? I = (0) 2 + (d) 2 = d 2 EXAMPLE 2: Consider two asses joined by a asless rod. d
a) What is the oentu of inertia with respect to an axis passing through one end? I CM = (d/2) 2 + (d/2) 2 = d 2 / 2 d b) What is the oentu of inertia with respect to an axis passing through the center of ass? I CM = d 2 d EXAMPLE 3: Consider five asses joined by a asless circular rod. What is the oentu of inertia with respect to an axis passing through the center of ass? I CM = 5 R 2 R
In the exaple above, what would be the values of I CM if the ring joining the five asses were not assless, but had a total ass M uniforly distributed? M R Answer: I CM = (M + 5) R 2. EXAMPLE 4: Consider of a ring of ass M, and radius R. What is the oentu of inertia with respect to an axis that passes through the CM and that is perpendicular to the plane of the ring? Assue the ass M is uniforly distributed. R M I CM = M R 2 EXAMPLE 5: Consider of a ring of ass M, and radius R. What is the oentu of inertia with respect to an axis that passes through the CM and that is in the plane of the ring. Assue the ass M is uniforly distributed..
I = i i r i 2 R r i M The suation turns out to be = (1/2)M R 2 EXAMPLE 6: Consider a thin bar of length d and total ass M (uniforly distributed). M d What is the oentu of inertia with respect to an axis that passes through the center of ass and that is perpendicular to the rod? 2 I = (d) i r i i I = L/2 r i (d) i M (d) L /2 r2 d = dr L/2 The suation turns out to be = ( dr) L /2 r2 = ( r 3 L / 2 = ( ( L ( -L L / 2 = ( ( L = ( ( L = ( L ( L L
Exaple: Find an expression for the total kinetic energy of a ring of radius R and ass M rolling down an inclined plane M V CM K = K =
But, what is the relationship between V CM and ω?
Helpful scheatic to figure out probles involving rotation without slipping. What is the relationship between: the speed v at which the center-point O advances, and the angular speed ω at which the cylinder of radius R rotates? Translational otion x O v B' ω s = R θ R O ds = R ω θ A dt s s A' = R α Rotational otion 1 If there is not slip then, x = s 2 Translational displaceent of point "O" Related to the rotational otion
x = s v = R ω Relationship between translational and rotational otion
Let's go back to our original question: Find an expression for the kinetic energy of a ring of radius R and ass M, rolling an inclined plane.