PHYS 1443 – Section 001 Lecture #13

PHYS 1443 – Section 001 Lecture #13
Thursday, June 22, 2006 Dr. Jaehoon Yu CM and the Center of Gravity Fundamentals on Rotational Motion Rotational Kinematics Relationship between angular and linear quantities Rolling Motion of a Rigid Body Torque Torque and Vector Product Moment of Inertia Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Announcements Reading assignments Last quiz next Wednesday
CH. 11.6, 11.8, 11.9 and 11.10 Last quiz next Wednesday Early in the class Covers Ch. 10 – what we cover next Tuesday Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Center of Mass and Center of Gravity
CM Axis of symmetry The center of mass of any symmetric object lies on an axis of symmetry and on any plane of symmetry, if object’s mass is evenly distributed throughout the body. One can use gravity to locate CM. Hang the object by one point and draw a vertical line following a plum-bob. Hang the object by another point and do the same. The point where the two lines meet is the CM. How do you think you can determine the CM of objects that are not symmetric? Since a rigid object can be considered as a collection of small masses, one can see the total gravitational force exerted on the object as Center of Gravity Dmi Dmig The net effect of these small gravitational forces is equivalent to a single force acting on a point (Center of Gravity) with mass M. What does this equation tell you? Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu The CoG is the point in an object as if all the gravitational force is acting on!

Motion of a Group of Particles
We’ve learned that the CM of a system can represent the motion of a system. Therefore, for an isolated system of many particles in which the total mass M is preserved, the velocity, total momentum, acceleration of the system are Velocity of the system Total Momentum of the system Acceleration of the system External force exerting on the system What about the internal forces? System’s momentum is conserved. If net external force is 0 Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Fundamentals on Rotation
Linear motions can be described as the motion of the center of mass with all the mass of the object concentrated on it. Is this still true for rotational motions? No, because different parts of the object have different linear velocities and accelerations. Consider a motion of a rigid body – an object that does not change its shape – rotating about the axis protruding out of the slide. The arc length is Therefore the angle, q, is And the unit of the angle is in radian. It is dimensionless!! One radian is the angle swept by an arc length equal to the radius of the arc. Since the circumference of a circle is 2pr, The relationship between radian and degrees is Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example 10 – 1 (a) One radian is 360o/2p. Thus
A particular bird’s eyes can barely distinguish objects that subtend an angle no smaller than about 3x10-4 rad. (a) How many degrees is this? (b) How small an object can the bird just distinguish when flying at a height of 100m? (a) One radian is 360o/2p. Thus (b) Since l=rq and for small angle arc length is approximately the same as the chord length. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Rotational Kinematics
The first type of motion we have learned in linear kinematics was under a constant acceleration. We will learn about the rotational motion under constant angular acceleration, because these are the simplest motions in both cases. Just like the case in linear motion, one can obtain Angular Speed under constant angular acceleration: Angular displacement under constant angular acceleration: One can also obtain Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Angular Displacement, Velocity, and Acceleration
Using what we have learned in the previous slide, how would you define the angular displacement? How about the average angular speed? qi qf Unit? rad/s And the instantaneous angular speed? Unit? rad/s By the same token, the average angular acceleration Unit? rad/s2 And the instantaneous angular acceleration? Unit? rad/s2 When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and angular acceleration. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Rotational Kinematics
A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, a) through what angle does the wheel rotate in 2.00s? Using the angular displacement formula in the previous slide, one gets Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Rotational Kinematics cnt’d
What is the angular speed at t=2.00s? Using the angular speed and acceleration relationship Find the angle through which the wheel rotates between t=2.00 s and t=3.00 s. Using the angular kinematic formula At t=2.00s At t=3.00s Angular displacement Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Relationship Between Angular and Linear Quantities
What do we know about a rigid object that rotates about a fixed axis of rotation? Every particle (or masslet) in the object moves in a circle centered at the axis of rotation. When a point rotates, it has both the linear and angular components in its motion. What is the linear component of the motion you see? The direction of w follows a right-hand rule. Linear velocity along the tangential direction. How do we related this linear component of the motion with angular component? The arc-length is So the tangential speed v is What does this relationship tell you about the tangential speed of the points in the object and their angular speed?: Although every particle in the object has the same angular speed, its tangential speed differs proportional to its distance from the axis of rotation. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu The farther away the particle is from the center of rotation, the higher the tangential speed.

Is the lion faster than the horse?
A rotating carousel has one child sitting on a horse near the outer edge and another child on a lion halfway out from the center. (a) Which child has the greater linear speed? (b) Which child has the greater angular speed? Linear speed is the distance traveled divided by the time interval. So the child sitting at the outer edge travels more distance within the given time than the child sitting closer to the center. Thus, the horse is faster than the lion. (b) Angular speed is the angle traveled divided by the time interval. The angle both the children travel in the given time interval is the same. Thus, both the horse and the lion have the same angular speed. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

How about the acceleration?
How many different linear accelerations do you see in a circular motion and what are they? Two Tangential, at, and the radial acceleration, ar. Since the tangential speed v is The magnitude of tangential acceleration at is Although every particle in the object has the same angular acceleration, its tangential acceleration differs proportional to its distance from the axis of rotation. What does this relationship tell you? The radial or centripetal acceleration ar is What does this tell you? The father away the particle is from the rotation axis, the more radial acceleration it receives. In other words, it receives more centripetal force. Total linear acceleration is Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example (a) What is the linear speed of a child seated 1.2m from the center of a steadily rotating merry-go-around that makes one complete revolution in 4.0s? (b) What is her total linear acceleration? First, figure out what the angular speed of the merry-go-around is. Using the formula for linear speed Since the angular speed is constant, there is no angular acceleration. Tangential acceleration is Radial acceleration is Thus the total acceleration is Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Rotational Motion
Audio information on compact discs are transmitted digitally through the readout system consisting of laser and lenses. The digital information on the disc are stored by the pits and flat areas on the track. Since the speed of readout system is constant, it reads out the same number of pits and flats in the same time interval. In other words, the linear speed is the same no matter which track is played. a) Assuming the linear speed is 1.3 m/s, find the angular speed of the disc in revolutions per minute when the inner most (r=23mm) and outer most tracks (r=58mm) are read. Using the relationship between angular and tangential speed b) The maximum playing time of a standard music CD is 74 minutes and 33 seconds. How many revolutions does the disk make during that time? c) What is the total length of the track past through the readout mechanism? d) What is the angular acceleration of the CD over the 4473s time interval, assuming constant a? Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Rolling Motion of a Rigid Body
What is a rolling motion? A more generalized case of a motion where the rotational axis moves together with the object A rotational motion about the moving axis To simplify the discussion, let’s make a few assumptions Limit our discussion on very symmetric objects, such as cylinders, spheres, etc The object rolls on a flat surface Let’s consider a cylinder rolling without slipping on a flat surface Under what condition does this “Pure Rolling” happen? The total linear distance the CM of the cylinder moved is R q s Thus the linear speed of the CM is s=Rq Condition for “Pure Rolling” Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

More Rolling Motion of a Rigid Body
The magnitude of the linear acceleration of the CM is As we learned in the rotational motion, all points in a rigid body moves at the same angular speed but at a different linear speed. P P’ CM vCM 2vCM CM is moving at the same speed at all times. At any given time, the point that comes to P has 0 linear speed while the point at P’ has twice the speed of CM Why?? A rolling motion can be interpreted as the sum of Translation and Rotation P P’ CM vCM P P’ CM v=Rw v=0 P P’ CM 2vCM vCM = + Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Torque Torque is the tendency of a force to rotate an object about an axis. Torque, t, is a vector quantity. F Consider an object pivoting about the point P by the force F being exerted at a distance r. f F2 r The line that extends out of the tail of the force vector is called the line of action. Line of Action P d2 d Moment arm The perpendicular distance from the pivoting point P to the line of action is called Moment arm. Magnitude of torque is defined as the product of the force exerted on the object to rotate it and the moment arm. When there are more than one force being exerted on certain points of the object, one can sum up the torque generated by each force vectorially. The convention for sign of the torque is positive if rotation is in counter-clockwise and negative if clockwise. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Torque A one piece cylinder is shaped as in the figure with core section protruding from the larger drum. The cylinder is free to rotate around the central axis shown in the picture. A rope wrapped around the drum whose radius is R1 exerts force F1 to the right on the cylinder, and another force exerts F2 on the core whose radius is R2 downward on the cylinder. A) What is the net torque acting on the cylinder about the rotation axis? F1 The torque due to F1 and due to F2 R1 R2 So the total torque acting on the system by the forces is F2 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 the rest? Using the above result The cylinder rotates in counter-clockwise. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Torque and Vector Product
x y z O Let’s consider a disk fixed onto the origin O and the force F exerts on the point p. What happens? t=rxF The disk will start rotating counter clockwise about the Z axis r p The magnitude of torque given to the disk by the force F is q F But torque is a vector quantity, what is the direction? How is torque expressed mathematically? What is the direction? The direction of the torque follows the right-hand rule!! The above operation is called Vector product or Cross product What is the result of a vector product? What is another vector operation we’ve learned? Another vector Scalar product Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu Result? A scalar

Properties of Vector Product
Vector Product is Non-commutative What does this mean? If the order of operation changes the result changes Following the right-hand rule, the direction changes Vector Product of two parallel vectors is 0. Thus, If two vectors are perpendicular to each other Vector product follows distribution law The derivative of a Vector product with respect to a scalar variable is Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

More Properties of Vector Product
The relationship between unit vectors, Vector product of two vectors can be expressed in the following determinant form Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Moment of Inertia Measure of resistance of an object to changes in its rotational motion. Equivalent to mass in linear motion. Rotational Inertia: For a group of particles For a rigid body What are the dimension and unit of Moment of Inertia? Determining Moment of Inertia is extremely important for computing equilibrium of a rigid body, such as a building. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Moment of Inertia
In a system of four small spheres as shown in the figure, assuming the radii are negligible and the rods connecting the particles are massless, compute the moment of inertia and the rotational kinetic energy when the system rotates about the y-axis at angular speed w. x y Since the rotation is about y axis, the moment of inertia about y axis, Iy, is M l m b O This is because the rotation is done about y axis, and the radii of the spheres are negligible. Why are some 0s? Thus, the rotational kinetic energy is Find the moment of inertia and rotational kinetic energy when the system rotates on the x-y plane about the z-axis that goes through the origin O. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Calculation of Moments of Inertia
Moments of inertia for large objects can be computed, if we assume the object consists of small volume elements with mass, Dmi. The moment of inertia for the large rigid object is It is sometimes easier to compute moments of inertia in terms of volume of the elements rather than their mass How can we do this? Using the volume density, r, replace dm in the above equation with dV. The moments of inertia becomes Example: Find the moment of inertia of a uniform hoop of mass M and radius R about an axis perpendicular to the plane of the hoop and passing through its center. x y The moment of inertia is R O dm The moment of inertia for this object is the same as that of a point of mass M at the distance R. What do you notice from this result? Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Rigid Body Moment of Inertia
Calculate the moment of inertia of a uniform rigid rod of length L and mass M about an axis perpendicular to the rod and passing through its center of mass. The line density of the rod is x y so the masslet is x dx The moment of inertia is L What is the moment of inertia when the rotational axis is at one end of the rod. Will this be the same as the above. Why or why not? Since the moment of inertia is resistance to motion, it makes perfect sense for it to be harder to move when it is rotating about the axis at one end. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Torque & Angular Acceleration
Let’s consider a point object with mass m rotating on a circle. Ft m r What forces do you see in this motion? Fr The tangential force Ft and radial force Fr The tangential force Ft is The torque due to tangential force Ft is What do you see from the above relationship? What does this mean? Torque acting on a particle is proportional to the angular acceleration. What law do you see from this relationship? Analogs to Newton’s 2nd law of motion in rotation. How about a rigid object? The external tangential force dFt is r dFt dm O The torque due to tangential force Ft is The total torque is What is the contribution due to radial force and why? Contribution from radial force is 0, because its line of action passes through the pivoting point, making the moment arm 0. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Torque and Angular Acceleration
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. The rod is released from rest in the horizontal position. What are the initial angular acceleration of the rod and the initial linear acceleration of its right end? The only force generating torque is the gravitational force Mg L/2 Mg Since the moment of inertia of the rod when it rotates about one end We obtain Using the relationship between tangential and angular acceleration What does this mean? The tip of the rod falls faster than an object undergoing a free fall. Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Rotational Kinetic Energy
ri mi q O x y vi What do you think the kinetic energy of a rigid object that is undergoing a circular motion is? Kinetic energy of a masslet, mi, moving at a tangential speed, vi, is Since a rigid body is a collection of masslets, the total kinetic energy of the rigid object is Since moment of Inertia, I, is defined as The above expression is simplified as Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Total Kinetic Energy of a Rolling Body
What do you think the total kinetic energy of the rolling cylinder is? Since it is a rotational motion about the point P, we can write the total kinetic energy Where, IP, is the moment of inertia about the point P. P P’ CM vCM 2vCM Using the parallel axis theorem, we can rewrite Rotational kinetic energy about the CM Translational Kinetic energy of the CM Since vCM=Rw, the above relationship can be rewritten as What does this equation mean? Total kinetic energy of a rolling motion is the sum of the rotational kinetic energy about the CM And the translational kinetic of the CM Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Kinetic Energy of a Rolling Sphere
x h q vCM w Let’s consider a sphere with radius R rolling down a hill without slipping. Since vCM=Rw What is the speed of the CM in terms of known quantities and how do you find this out? Since the kinetic energy at the bottom of the hill must be equal to the potential energy at the top of the hill Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

Example for Rolling Kinetic Energy
For solid sphere as shown in the figure, calculate the linear speed of the CM at the bottom of the hill and the magnitude of linear acceleration of the CM. Solve this problem using Newton’s second law, the dynamic method. What are the forces involved in this motion? n f x y Gravitational Force, Frictional Force, Normal Force M x h q Mg Newton’s second law applied to the CM gives Since the forces Mg and n go through the CM, their moment arm is 0 and do not contribute to torque, while the static friction f causes torque We know that We obtain Substituting f in dynamic equations Thursday, June 22, 2006 PHYS , Summer 2006 Dr. Jaehoon Yu

PHYS 1443 – Section 001 Lecture #13

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PHYS 1443 – Section 001 Lecture #13

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