Chapter 10: Rotational Motion & Angular Momentum PDF

chapter10: rotational motion & angular momentum Mrs. Pooja Brahmaiahchari Angular Acceleration In uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. The angular velocity was defined as the time rate of change of a...

chapter10: rotational motion & angular momentum Mrs. Pooja Brahmaiahchari Angular Acceleration In uniform circular motion, which is motion in a circle at constant speed and, hence, constant angular velocity. The angular velocity was defined as the time rate of change of angle θ: ∆𝜃 𝜔= ∆𝑡 The relationship between angular velocity and linear velocity was also defined in Rotation Angle and Angular Velocity as, 𝑣 = 𝑟𝜔 Or 𝑣 𝜔= 𝑟  According to the sign convention, the counter clockwise direction is considered as positive direction and clockwise direction as negative. Angular velocity is not constant when a skater pulls in her arms, when a child starts up a merry-go-round from rest, or when a computer’s hard disk slows to a halt when switched off. In all these cases, there is an angular acceleration, in which ω changes. The faster the change occurs, the greater the angular acceleration.  Angular acceleration α is defined as the rate of change of angular velocity. In equation form, angular acceleration is expressed as follows: ∆𝜔 𝛼= ∆𝑡 where ∆𝜔 is the change in angular velocity and ∆𝑡 is the change in time. The units of angular acceleration are rad/s2, or If 𝝎 increases, then 𝜶 is positive. If 𝝎 decreases, then 𝜶 is negative. 1) Suppose a girl puts her bicycle on its back and starts the rear wheel spinning from rest to a final angular velocity of 250 rpm in 5.00s. Calculate the angular acceleration in rad/s2. 2) At its peak, a tornado is 60.0 m in diameter and carries 500 km/h winds. What is its angular velocity in revolutions per second?  If the bicycle in the preceding example had been on its wheels instead of upside-down, it would first have accelerated along the ground and then come to a stop. This connection between circular motion and linear motion needs to be explored. In circular motion, linear acceleration is tangent to the circle at the point of interest, as seen in Figure. Thus, linear acceleration is called tangential acceleration.  Linear or tangential acceleration refers to changes in the magnitude of velocity but not its direction. That in circular motion centripetal acceleration, ac, refers to changes in the direction of the velocity but not its magnitude. An object undergoing circular motion experiences centripetal acceleration, Thus, at and ac are perpendicular and independent of one another. Tangential acceleration at is directly related to the angular acceleration and is linked to an increase or decrease in the velocity, but not its direction.  The relationship between linear acceleration at and angular acceleration α. Because linear acceleration is proportional to a change in the magnitude of the velocity, it is defined to be ∆𝑣 𝑎𝑡 = ∆𝑡 ∆𝑟𝜔 𝑟∆𝜔 For circular motion, v = rω, 𝑎𝑡 = 𝑜𝑟 𝑎𝑡 = ∆𝑡 ∆𝑡 ∆𝜔 By definition, 𝛼 = Thus, 𝑎𝑡 = 𝑟𝛼 ∆𝑡 𝑎𝑡 Or 𝛼= 𝑟 These equations mean that linear acceleration and angular acceleration are directly proportional. The greater the angular acceleration is, the larger the linear (tangential) acceleration is, and vice versa.  So far, we have defined three rotational quantities—θ, ω, α. These quantities are analogous to the translational quantities x,v, a. The table given below describes the relationship between them. Kinematics of Rotational Motion The kinematics of rotational motion describes the relationships among rotation angle, angular velocity, angular acceleration, and time. 3. A deep-sea fisherman hooks a big fish that swims away from the boat pulling the fishing line from his fishing reel. The whole system is initially at rest and the fishing line unwinds from the reel at a radius of 4.50 cm from its axis of rotation. The reel is given an angular acceleration of 110 rad/s2 for 2.00 s as seen in figure. (a) What is the final angular velocity of the reel? (b) At what speed is fishing line leaving the reel after 2.00 s elapses? (c) How many revolutions does the reel make? 4. A gyroscope slows from an initial rate of 32.0 rad/s at a rate of 0.700 rad/s2 (a) How long does it take to come to rest? (b) How many revolutions does it make before stopping? 5. Suppose a yo-yo has a center shaft that has a 0.250 cm radius and that its string is being pulled. (a) If the string is stationary and the yo-yo accelerates away from it at a rate of 1.50 m/s^2 , what is the angular acceleration of the yo-yo? (b) What is the angular velocity after 0.750 s if it starts from rest? (c) The outside radius of the yo-yo is 3.50 cm. What is the tangential acceleration of a point on its edge? Dynamics of Rotational Motion: Rotational Inertia To develop the precise relationship among force, mass, radius, and angular acceleration, consider what happens if we exert a force on a point mass that is at a distance from a pivot point, as shown in Figure. Because the force is perpendicular to r, an acceleration a = F/m is obtained in the direction of F. Rearranging the equation we get F = ma  We know that a = rα.We get : 𝐹 = 𝑚𝑟𝛼 Recall torque is the turning effectiveness of a force. In this case, because F is perpendicular to r, torque is simply 𝜏 = 𝐹𝑟. Multiplying both sides by r we get, 𝑟𝐹 = 𝑚𝑟 2 𝛼 𝑜𝑟 𝜏 = 𝑚𝑟 2 𝛼 This last equation is the rotational analog of Newton’s second law (F = ma), where torque is analogous to force, angular acceleration is analogous to translational acceleration, and mr2 is analogous to mass (or inertia). The quantity mr2 is called the rotational inertia or moment of inertia of a point mass a distance from the center of rotation. Rotational Inertia and Moment of Inertia we define the moment of inertia I of an object to be the sum of mr2 for all the point masses of which it is composed. That is, 𝐼 = σ 𝑚𝑟 2 Here I is analogous to m in translational motion. Because of the distance r, the moment of inertia for any object depends on the chosen axis. The general relationship among torque, moment of inertia, and angular acceleration is 𝑛𝑒𝑡 𝜏 = 𝐼𝛼 𝑜𝑟 𝑛𝑒𝑡 𝜏 𝛼= 𝐼 6. Consider the father pushing a playground merry-go-round in figure. He exerts a force of 250 N at the edge of the 50.0-kg merry-go-round, which has a 1.50 m radius. Calculate the angular acceleration produced (a) when no one is on the merry-go round and (b) when an 18.0-kg child sits 1.25 m away from the center. Consider the merry-go-round itself to be a uniform disk with negligible retarding friction. > - I = "2mr a) = Y F , I= +Mr = Rotational Kinetic Energy: Work and Energy Revisited Work must be done to rotate objects such as grindstones or merry-go- rounds. The simplest rotational situation is one in which the net force is exerted perpendicular to the radius of a disk and remains perpendicular as the disk starts to rotate. The equation, 𝑛𝑒𝑡 𝑊 = 𝑛𝑒𝑡 𝜏 𝜃 is expression for rotational work.  To get an expression for rotational kinetic energy, we must again perform some algebraic manipulations. The first step is to note that 𝑛𝑒𝑡 𝜏 = 𝐼𝛼, so that , 𝑛𝑒𝑡 𝑊 = 𝐼𝛼𝜃 we solve one of the rotational kinematics equations for 𝛼𝜃. We start with the equation 𝜔2 = 𝜔02 + 2𝛼𝜃 Solving for 𝛼𝜃 we get, 𝜔2 − 𝜔02 𝛼𝜃 = 2  The work-energy theorem for rotational motion only equation is: 𝟏 𝟐 𝟏 𝒏𝒆𝒕 𝑾 = 𝑰𝝎 − 𝑰𝝎𝟐𝟎 𝟐 𝟐 Through an analogy with translational motion, we define the term 1 𝐼𝜔2 to be rotational kinetic energy for an object with a moment of 2 inertia and an angular velocity ω: 1 𝐾𝐸𝑟𝑜𝑡 = 𝐼 𝜔2 2 Why Don’t All Objects Roll Downhill at the Same Rate? One of the quality controls in a tomato soup factory consists of rolling filled cans down a ramp. If they roll too fast, the soup is too thin. Why should cans of identical size and mass roll down an incline at different rates? And why should the thickest soup roll the slowest? Suppose each can starts down the ramp from rest. Each can starting from rest means each starts with the same gravitational potential energy 𝑃𝐸𝑔𝑟𝑎𝑣 , which is converted entirely to KE, provided each rolls without slipping. KE however, can take the form of 𝐾𝐸𝑡𝑟𝑎𝑛𝑠 or 𝐾𝐸𝑟𝑜𝑡 , and total 𝐾𝐸 is the sum of the two.  If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. Thus, the can goes slower than it would if it slid down. Furthermore, the thin soup does not rotate, whereas the thick soup does, because it sticks to the can. The thick soup thus puts more of the can’s original gravitational potential energy into rotation than the thin soup, and the can rolls more slowly, as seen in Figure  Assuming no losses due to friction, there is only one force doing work—gravity. Therefore, the total work done is the change in kinetic energy. As the cans start moving, the potential energy is changing into kinetic energy. Conservation of energy gives, 𝑃𝐸𝑖 = 𝐾𝐸𝑓 More specifically, 𝑃𝐸𝑔𝑟𝑎𝑣 = 𝐾𝐸𝑡𝑟𝑎𝑛𝑠 + 𝐾𝐸𝑟𝑜𝑡 Or 1 1 𝑚𝑔ℎ = 𝑚 𝑣 + 𝐼𝜔2 2 2 2 7. What is the final velocity of the hoop that rolls without slipping down a 5.00m high hill, starting from the rest? Angular Momentum and Its Conservation By now the pattern is clear—every rotational phenomenon has a direct translational analog. It seems quite reasonable, then, to define angular momentum L as 𝐿 = 𝐼𝜔 This equation is an analog to the definition of linear momentum as p = mv. Units for linear momentum are kg. m/s. while units for angular momentum are kg. m2/s. When you push a merry-go-round, spin a bike wheel, or open a door, you exert a torque.  If the torque you exert is greater than opposing torques, then the rotation accelerates, and angular momentum increases. The greater the net torque, the more rapid the increase in L. The relationship between torque and angular momentum is ∆𝐿 𝑛𝑒𝑡 𝜏 = ∆𝑡 It is in fact the rotational form of Newtons second law. Conservation of Angular momentum This equation ∆𝐿 = (𝑛𝑒𝑡 𝜏)∆𝑡 means that, to change angular momentum, a torque must act over some period of time. Because Earth has a large angular momentum, a large torque acting over a long time is needed to change its rate of spin. So, what external torques are there? Tidal friction exerts torque that is slowing Earth’s rotation, but tens of millions of years must pass before the change is very significant. If the net torque is zero, then angular momentum is constant or conserved. ∆𝐿 We can see this rigorously by considering 𝑛𝑒𝑡 𝜏 = for the situation in ∆𝑡 which the net torque is zero.  In that case, 𝑛𝑒𝑡 𝜏 = 0 ∆𝐿 Implying that = 0, ∆𝑡 If the change in angular momentum ∆L is zero, then the angular momentum is constant; thus, 𝐿 = 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡 (𝑛𝑒𝑡 𝜏 = 0) Or 𝐿 = 𝐿′ (𝑛𝑒𝑡𝜏 = 0) 8. The triceps muscle in the back of upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2x103 N with an effective perpendicular lever arm of 3cm, producing an angular acceleration of the forearm of 120 rad/s2. What is Moment of Inertia of boxer’s forearm? 9. A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her leg. She produces angular acceleration of 30 rad/s2 and her lower leg has a moment of Inertia 0.750 kg m2. What is the force exerted by the muscle if its effective perpendicular lever arm is 1.90cm? 10. While punting a football, kicker rotates his leg about the hip joint. The moment of Inertia of leg is 3.75 kgm2 and its rotational kinetic energy is 175 J. a) What is angular velocity of leg. B) What is the velocity of tip of punter’s shoe if it is 1.05m from hip joint? THANK YOU