Presentation on theme: "A 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg, placed (x) m from the support point of a uniform beam. What."— Presentation transcript:
A 40-kg mass placed 1.25 m on the opposite side of the support point balances a mass of 25 kg, placed (x) m from the support point of a uniform beam. What is the value of the unknown distance x? Example
A large seed initially at rest explodes into two pieces which move off. Which of these could be possible paths the two pieces would take? (I) (II) (III)
A ball is projected straight up. Which graph shows the total energy of the ball as a function of time? t tt t t t (A) (B) (C) (D) (E) (F)
Rotational Inertia (I) Recall: From Newton’s 2 nd law of linear motion: F net = ma = m(v-v o )/ t mass (m) = measure of inertia of an object. = measure of how difficult it is to = measure of how difficult it is to change linear velocity (v) of an object. change linear velocity (v) of an object. The larger the mass, the more difficult it is to change its velocity v. For rotation, moment of inertial or rotational inertia (I) = measure of how difficult it is to change angular velocity ( ) of an object.
Rotational Inertia (I) Rotational inertia (I) = measure of how difficult it is to change angular velocity ( ) of an object. Rotational inertia (I) – Also called moment of inertia: Depends on mass, m. (I m). Depends on mass, m. (I m). Depends on square of radius of rotation (I r 2 )Depends on square of radius of rotation (I r 2 ) Depends on how mass m is distributed. Depends on how mass m is distributed. Same mass. Different radius. Which one easier to get to start rotating?
Rotational Inertia For a point object, I = mr 2 Point object = one whose size is small c.f. radius r r m Units of I = kg-m 2
Rotational Inertia For a point object, I = mr 2 r1r1 m1m1 For many discrete point objects with different shapes rotated about the same axis, Total rotational inertia = sum of I for each object. I total = m i r i 2
Rotational Inertia l For a point object, I = mr 2 l For objects of comparable size to radius r, the moment of in inertia depends on distribution of mass.
Rotational Inertia (I) Mass of earth = 5.975 x 10 24 kg Mean radius = 6.37 x 10 6 m Earth-sun distance = 1.5 10 11 m What is the rotational inertia of the earth’s spin about its axis?
Example Two solid spherical balls are joined together by a 4-meter long steel rod. If they are spun about a vertical axis passing through their center of mass, find the total rotational inertia. [Hint, treat the spinning balls as point objects with I = mR 2 ]. m 1 = 3 kg m 2 = 5 kg
Rotational Inertia (I) LinearAngular (Rotational) Displacement (x) Angle ( ) in radians Velocity (v) Angular Velocity ( ) Acceleration (a) Angular Acceleration ( ) Mass (m)Rotational Inertia (I) K tran = ½ mv 2 K rot = ½ I 2
Rotational Kinematics LinearRotational a = constant = constant v = v o + at = o + t x = x o + v o t + ½at 2 = o + o t + ½ t 2 v 2 = v o 2 + 2a(x - x o ) 2 = o 2 + 2 ( - o )
Comment on axes and sign (i.e. what is positive and negative) Whenever we talk about rotation, it is implied that there is a rotation “axis”. This is usually called the “z” axis (we usually omit the z subscript for simplicity). Counter-clockwise (increasing ) is usually called positive. Clockwise (decreasing ) is usually called negative. z
Example You and a friend are playing on a merry-go- round. You stand at the outer edge of the merry-go-round and your friend stands halfway between the outer edge and the center. Assume the rotation rate of the merry- go-round is constant. Who has the greatest angular velocity? 1. You do 2. Your friend does 3. Same
Example Who has the greatest tangential velocity (v)? 1. You do 2. Your friend does 3. Same
Rotational Kinetic Energy m r v Mass m in rotational motion. Its rotational inertia, I = mr 2 Since it is moving, it has kinetic energy. K = ½ mv 2 From v = r , K = ½ mr 2 2 = ½ (mr 2 ) 2 = ½I 2
Kinetic Energy of Rotating Disk Consider a disk with radius R and mass M, spinning with angular frequency Each “piece” of disk has speed v i = r i è Each “piece” has kinetic energy »K i = ½ m i v 2 » = ½ m i 2 r i 2 è Combine all the pieces » K i = ½ m i 2 r i 2 » = ½ ( m i r i 2 ) 2 » = ½ I 2 riri mimi
Rotational Kinetic Energy (K rot ) A rigid object spinning about a fixed axis (pure rotational motion) has rotational kinetic energy K rot = ½I 2 [eg, spinning wheel] If the rigid object is moving (sliding) with velocity v without any spin (ie pure translational motion), it has only translational kinetic energy K = ½ mv 2 [eg, skidding wheel]
Rotational Kinetic Energy (K rot ) If the rigid object is spinning with angular velocity while its center of mass moves linearly with velocity v cm, it has both translational and rotational kinetic energy. Its total kinetic energy will be K tot = K tran + K rot ½ mv cm 2 + ½ I 2 [eg, wheel rolling normally on the ground, ball rolling on the ground]
è K tran = ½ m v 2 Linear Motion K rot = ½ I 2 Rotational Motion
Example Who has the greater kinetic energy? (Assume masses are equal) 1. You do 2. Your friend does 3. Same
Example A 10-kg hollow cylindrical shell rolls on the ground at a linear velocity of 5 m/s. Find its total kinetic energy. 50 cm
Torque ( l Force (F) is responsible for change of linear velocity. Net force results in linear acceleration. Torque ( is responsible for change in angular velocity. Net torque results in angular acceleration.
Torque r = r F = (rsin )F = rFsin r = rsin = lever arm (moment arm) Torque = lever arm x force Unit: m-N F rr
Larger lever arm r , larger torque if F stays unchanged Larger force F, larger torque if lever arm stays unchanged 3. If r = 0, If = 0, If = 90 o, rF F r = r F = rsin F
Torque Counter clockwise rotation, torque is positive F F Clockwise rotation, torque is negative Torque is a vector quantity.
A string is tied to a doorknob 0.80 m from the hinge as shown in the figure. At the instant shown, the force applied to the string is 5.0 N. What is the torque on the door? hinge
Equilibrium l Recall, translational equilibrium – F net = 0. l F net = 0, object at rest – static equilibrium. l F net = 0, object moving with constant velocity, - dynamic equilibrium. l Rotational equilibrium, net = 0. l An object in equilibrium means it is in both translational equilibrium and rotational equilibrium. l Conditions for equilibrium: F net = 0 and net = 0
Equilibrium Conditions for equilibrium: F net = 0 and net = 0 l Torque can be calculated about any desired axis. l A judicious choice of axis helps. l Choose axis at a point through which an unknown force acts so that its torque does not appear in the equation.
l Is the point through which the force of gravity acts. l Essentially the same as center of mass. Center of Gravity
1.5 m 2.5 m M 1 = 50 kg M 2 = 60 kg pivot The figure above is a snapshot of two masses on a light beam placed on a pivot. What is the net torque about the pivot? To be in rotational equilibrium, what should have been the value of M 2 ?
Example The picture below shows two people lifting a heavy log. Which of the two people is supporting the greatest weight? 1. The person on the left is supporting the greatest weight 2. The person on the right is supporting the greatest weight 3. They are supporting the same weight
Example The picture below shows two people (L & R) lifting a 20-kg log. The length of the log is 4.0 m and L is 1.0 m from the left end while R is holding from the right end. Find how much force each of them have to exert. L R
l Newton’s 2 nd law for linear motion: F net = ma Eg. What force is required to change velocity of a 6 kg solid spherical object of radius 30 cm object from 15 m/s to 25 m/s in 5 seconds? l Newton’s 2 nd law for rotation: net = I Eg. What torque is required to change angular velocity of a 6 kg solid spherical object of radius 30 cm object from 15 rad/s to 25 rad/s in 5 seconds? [Assume spin about the axis] Newton’s 2 nd Law
Work Done by Torque l Recall, work done by constant force W = F x cos Also, W net = K = ½ mv f 2 – ½ mv i 2 l For rotation, work done by constant torque: W = F tangential s = F tangential r = Also, W net = K rot = ½ I f 2 – ½ I i 2 Power P = W/t = /t
Work Done by Torque A 182-kg flywheel has an effective radius of 0.62 m. (a)How much torque will take it from rest to a rotational speed of 120 rpm within 30.0 s? (b) How much work will this take?
l A rolling object posses both K tran (½ mv 2 CM ) and K rot (½ I 2 ) Rolling Object E 1 = U 1 + K 1 = mgh + 0 E 2 = U 2 + K 2 0 + ½ mv 2 + ½ I 2 h If it came down rolling: mgh = ½ mv A 2 + ½ I 2 If it came down only skidding, mgh = ½ mv B 2
Angular Momentum (L) Recall: Linear Momentum: p = mv Angular Momentum, L = I Angular momentum of a rigid object rotating about a fixed axis. L is a vector quantity. Units = kg-m 2 /s Direction: If rotation is CCW, L is upward. If rotation is CW, L is downward.
Conservation of Angular Momentum If net external torque acting on a rotating object is zero, total angular momentum will be conserved. If net = 0, L before = L after OR I i i = I f f EG: Skater spinning with her arms extended away from her body. L before = I i i = (mr i 2 ) i She then decides to pull her arms close to her body so that her spinning radius becomes smaller (r f < r i ) and L after = I f f = (mr f 2 ) f
Conservation of Angular Momentum Since no external torque acts on her, principle of conservation of angular momentum will hold: L before = L after or I i i = I f f or (mr i 2 ) i = (mr f 2 ) f Since r i > r f, i < f ie, the skater will spin faster by just folding her arms.
Example A 1000-kg merry-go-round of radius 8.0 m is spinning at a uniform speed of 25 rpm. If a 150-kg mass is gently placed at the edge, what will be the final spin speed of the merry-go-round?
Example Which of the forces in the figure below produces the largest torque about the rotation axis indicated? Assume all the four forces have equal magnitudes. Axis 1 2 3 4 (A) 2 (B) 3 (C) 4 (D) The torques are equal
A 250-kg merry-go-round of radius 2.2 m is spinning at a uniform angular speed of 3.5 rad/s. Calculate the magnitude of its angular momentum. Use I = ½ MR 2 A.605 kg-m 2 /s B.1.93 x 10 3 kg-m 2 /s C.7.41 x 10 3 kg-m 2 /s D.2.12 x 10 3 kg-m 2 /s E.963 kg-m 2 /s
A 30.0-cm long wrench is used to generate a torque of 14.6 N-m at a bolt. A force of 70.0 N is applied at the end of the wrench at an angle of to the wrench. The angle is A.44 o B.0.4 o C.9.2 o D.3.6 o
A 3.2-kg solid sphere of diameter 28 cm spins at 5.0 rad/s about an axis passing through its center. How much work is needed to bring it to rest within 15 sec? [Use I = 2/5 x mr 2 ] A.0.025 J B.1.25 J C.1.25x10 4 J D.3.14 x 10 3 J E.0.31 J
Abel and Brian support a 68.0-kg uniform bar that is 4.0 m long. Abel holds the bar 1.0 m from one end, while Brian holds from the other end. How much force does Abel have to exert? A.444 N B.45.3 N C.1.33x10 3 N D.333 N E.222 N