Chapter 10 Rotational Kinematics Up to now we have only considered pointparticles i e we have not considered their shape or size only their mass Also we have only considered the motion of point particles straight line free fall projectile motion But real objects can also tumble twirl This subject rotation is what we explore in this chapter and in Chapter 11 First we begin by extending the concepts of circular motion Instead of a point particle consider a thin disk of radius r spinning on its axis y This disk is a real object it has structure r x We call these kinds of objects Rigid Bodies z Axis of rotation r r s arc length s Rigid Bodies do not bend twist or flex for example a billiard ball arc length s radius r Units of radians rad s r 2 rad s 2 r circumference For one complete revolution Conversion relation 2 rad 360 Now consider the rotation of the disk from some initial angle i to a final angle f during some time period ti to tf f i Angular displacement units of rad ccw is f i avg t f t i t Average angular velocity units of rad s y ccw f i x z Similar to instantaneous velocity we can define the Instantaneous Angular Velocity lim t 0 t A change in the Angular Velocity gives f i avg t f ti t Average Angular Acceleration rads s2 Analogous to Instantaneous Angular Velocity the Instantaneous Angular Acceleration is lim t 0 t Actually the Angular Velocities and Angular Acceleration are magnitudes of vector quantities and What is their direction They point along the axis of rotation with the sign determined by the right hand rule Example A fan takes 2 00 s to reach its operating angular speed of 10 0 rev s What is the average angular acceleration rad s2 Solution Given tf 2 00 s f 10 0 rev s Recognize ti 0 i 0 and that f needs to be converted to rad s f 10 0 rev s 2 rads 20 0 1 rev rad s 62 8 rads avg f i 20 0 0 10 0 t f ti 2 00 0 avg 31 4 rad s2 rad s2 Equations of Rotational Kinematics Just as we have derived a set of equations to describe linear or translational kinematics we can also obtain an analogous set of equations for rotational motion Consider correlation of variables Translational Rotational x displacement v velocity a acceleration t time t Replacing each of the translational variables in the translational kinematic equations by the rotational variables gives the set of rotational kinematic equations for constant 1 f i i f t f t i 2 2 1 f i i t f t i 2 t f t i f i t f t i 2 2 f i 2 f i We can use these equations in the same fashion we applied the translational kinematic equations Example Problem A figure skater is spinning with an angular velocity of 15 rad s She then comes to a stop over a brief period of time During this time her angular displacement is 5 1 rad Determine a her average angular acceleration and b the time during which she comes to rest Solution Given f 5 1 rad i 15 rad s Infer i 0 f 0 ti 0 Find tf a Use last kinematic equation 2 f 2 i 2 f i 2 0 i 2 f 2 2 i 15 rad s rad 22 s2 2 f 2 5 1 rad b Use first kinematic equation f i 12 i f t f ti 1 f 0 2 i 0 t f 0 2 f 2 5 1 rad tf 0 68 s i 15 rad s Or use the third kinematic equation f i t f t i 0 i t f i 15 rad s tf 0 68 s 2 22 rad s Example Problem At the local swimming hole a favorite trick is to run horizontally off a cliff that is 8 3 m above the water tuck into a ball and rotate on the way down to the water The average angular speed of rotation is 1 6 rev s Ignoring air resistance determine the number of revolutions while on the way down Solution y yi vi yf x Given i f 1 6 rev s yi 8 3 m Also vyi 0 ti 0 yf 0 Recognize two kinds of motion 2D projectile motion and rotational motion with constant angular velocity Method revolutions t Therefore need to find the time of the projectile motion tf Consider y component of projectile motion since we have no information about the xcomponent 1 2 y f yi v yi t f ti a y t f ti 2 2 2 1 1 0 yi 2 gt f yi 2 gt f 2 yi 2 8 3 m tf 1 3 s g 9 80 sm2 f i 12 i f t f ti t f f 1 6 rev s 1 3 s 2 1 rev Tangential Velocity r v For one complete revolution the angular displacement is 2 rad From Uniform Circular Motion the time for a complete revolution is a period T z Therefore the angular velocity frequency can be written 2 t T rad s Also the speed for an object in a circular path is 2 r v r v T T Tangential speed m s The tangential speed corresponds to the speed of a point on a rigid body a distance r from its center rotating at an angular speed Each point on the rigid body rotates at the same angular speed but its tangential speed depends on its location r vT r r 0 If the angular velocity changes is not constant then we have an angular acceleration For some point on a disk for example v t i r i v t f r f From the definition of translational acceleration a v f vi r f r i f i r t t t at r Tangential acceleration units of m s2 Since the speed changes this is not Uniform Circular Motion Also the Tangential Acceleration is different from the Centripetal Acceleration 2 Recall 2 t v v ac ar r r at We can find a total resultant acceleration a since at and ar are perpendicular 2 r 1 2 t a a a tan at ar Previously for the case of uniform circular motion at 0 and a ac ar The acceleration vector pointed to the center of the circle ar z at a ar If at 0 acceleration points away from the center Example A thin rigid rod is rotating with a constant angular acceleration about an axis that passes perpendicularly through one of its ends At one instant the total acceleration vector radial plus tangential at the other end of the rod makes a 60 0 angle with respect to the rod and has a magnitude of 15 0 m s2 The rod has an angular speed of 2 00 …
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