Chap10. Rotation of a Rigid Object about a Fixed Axis

Transcription

1 Chap10. Rotation of a Rigid Object about a Fixed Axis Level : AP Physics Teacher : Kim 10.1 Angular Displacement, Velocity, and Acceleration - A rigid object rotating about a fixed axis through O perpendicular to the plane. - An arbitrary point P is at a fixed distance r from the origin and rotates about it in a circle or radius r. - However, every particle on the object undergoes circular motion about O. -The relationship between the arc length s, radius r and angle θ is s = r θ θ = s r - θ is the ratio of an arc length and the radius of the circle with the unit radian, where one radian is the angle subtended by an arc length equal to the radius of the arc 1rad = 57.3 = 360 2π = 0.159revs - Angular displacement is defined as θ= θ f θ i [rad] - Average angular speed is defined as w = θ unit [rad/s] t - Instantaneous angular speed is defined as w = dθ dt - Instantaneous angular acceleration is defined as α = dw dt unit [rad/s2 ] compare with v= dx dt compare with a= dv dt When rotating about a fixed axis, every particle on a rigid object rotates through the same angle and has the same angular speed and the same angular acceleration - The directions of w and α For rotation about a fixed axis, the only direction that uniquely specifies the rotational motion is the direction along the axis of rotation. Therefore, the directions are of w and α are along this axis. If an object rotates in the xy plane, the direction of w is out of the plane when the rotation is counterclockwise Counterclockwise, w > 0, but if clockwise, then w < 0. See right-hand rule in p Fig.10.3 If w increases in time, then w & α is in same direction. If w decreases in time, then w & α are opposite

2 10.2 Analysis Model : Rigid Object Under Constant Angular Acceleration When an object s rotational motion is under constant angular acceleration, the kinematic relationships can be expressed as w f =w i + αt, θ f = θ i + w it αt2, w f 2 = w i 2 + 2α( θ f θ i ) - Review Example10.1, p.297, and solve problems #5, 7, 9 on p.325 p.297 Ex1 ) Rotating Wheel p.325 #5. #7. #9.

3 10.3 Angular and Translational Quantities - For a rotating body, there coexist the angular speed and tangential speed. - Point P moves in a circle where the linear velocity v is always tangent to the circular path. Hence, called tangential velocity. - The tangential speed and angular speed has a relationship as v t = rw - The tangential acceleration and angular acceleration has a relationship as at = dv dw =r dt dt = rα - Tangential acceleration will only exist if point P is moving with changing tangential speed - Furthermore, if an object is moving in circular motion, there is another acceleration, called centripetal acceleration (or radial acceleration) ar = v² r = rw² - The total acceleration will then be a = a t + a r (in vector form) - The magnitude of a for the point P is a = at² + ar² = r²α² + r²w 4 = r α² + w 4 - Solve problems #15, 17 on p.325 #15. #17.

4 10.4 Torque Review! Torque is the measure of the ability of a force to rotate an object around an axis. Torque is defined as τ = rfsinφ where τ is the torque, r is the distance between the axis and the position of the force applied and φ is the angle between the force and lever arm (or moment arm). Axis F φ - Torque will be maximum when φ=90. That is rotation of the object will occur best(easier) when a force is applied φ=90 - This is the reason why the doorknob is lever arm position as far away from the axis (hinge of or moment arm the door) as possible. r - We choose counter-clockwise as positive direction *~To unscrew a bolt, we need to rotate the bolt. Whenever you want to rotate any object, you must apply a torque on the object. 1. Calculate the torque produced by a 50N perpendicular force at the end of a 0.5m long wrench. a) 18N m b) 22N m c) 25N m d) 35N m 0.5m Push with 50N of force! Net Torque There can be more than one torque acting on an object. (recall net force=> F=ma) where τ = rfsinφ τ =τ 1 + τ 2 + τ n 2. A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meter-stick. If 10N of pushing force are applied as shown at the edge at both ends, find the net torque acting on the system. (and which direction? clockwise or counter-clockwise?) a) 10N m b) 15N m c) 20N m d) 25N m F=10N 0.5m axis 0.5m *~1m =100cm F=10N

5 3. A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meter-stick. If 15N and 10N of pushing force are applied as shown at the edge at both ends, find the net torque acting on the system. (and which direction? clockwise or counter-clockwise?) a) 1.0N m b) 1.5N m c) 2.0N m d) 2.5N m F=15N F=10N 4. A meter-stick is pivoted at the center. Any unbalance torque applied will result in rotating the meter-stick. If 15N, 10N and 5N of pushing force are applied as shown, find the net torque acting on the system. The 10N force is applied 0.25m away from the center(and which direction? clockwise or counterclockwise?) a) 1.0N m b) 1.5N m c) 2.0N m d) 2.5N m F=15N F=10N F=5N 5. A string is tied to a doorknob 0.79m from the hinge as shown(top-view of a door). The force applied to the string is 5N. What is the torque on the door? a) 3.3N m b) 2.2N m c) 1.1N m d) 0.8N m hinge 57 F Review Example10.3 p.301 The Net Torque on a Cylinder 10.5 Analysis Model : Rigid Object Under a Net Torque To make an object rotate, we must apply a torque(=force at a certain distance away from the axis) Στ = Iα where, torque is τ = rfsinφ, I is the moment of inertia is and α is angular acceleration Review Example 10.4 p.305 Rotating Rod

6 Solve #55. P.329 (a), (b), (c) only 10.6 Calculation of Moments of Inertia What is Moment of Inertia(or Rotational Inertia)? - Every rotating object has the tendency to keep its state of rotational motion - More massive the object, the greater the tendency to keep is state rotational of motion. However, the mass distribution also plays an important role. - Moment of inertia I is defined as I = lim Δm 0 ri² m i = r²dm - In the figure below, case-1 will greater moment of inertia not because the mass is greater but because the mass is located farther away from the axis than case-2. - Since r is larger in case-1, according to the formula I=Σmr 2, rotational inertia I is greater for case-1. Case-1 r r Case-2 r r - If you actually rotate the two objects, you will discover that case-1 will be harder to rotate than case-2 The greater the moment of inertia an object has, the greater the resistance to changes to state of rotation. That is, if an object has a lot of moment of inertia, it will harder to get it rotating, or if it is already rotating it will be harder to slow it down - Review example 10.7, p.308 Uniform Rigid Rod

7 - Review example 10.8, 309 Uniform Solid Cylinder - Parallel-axis theorem states that the moment of inertia of an object about any axis parallel to and a distance D away from this axis is I = I CM + MD 2 where I CM is the moment of inertia of the object when the axis is at the center, M is the total mass of the object and D is the distance from the actual axis to the location of the center of mass axis If the axis is located at a random position of the rod, then it COM! can be difficult to use integration to calculate the moment of inertia. D By using the parallel-axis theorem, we can calculate the momentum of inertia without integration and with a more simple calculation - Review Example 10.9, p.310 Applying the Parallel-Axis Theorem

8 10.7 Rotational Kinetic Energy - The rotational kinetic energy of a rotating rigid object is expressed as K R = 1 2 Iw2 See p.311for derivation - The moment of inertia I (or rotational inertia) is a measure of the resistance of an object to changes in its rotational motion, just as mass is a measure of the resistance of an object to changes in its linear motion. - Note that mass is an intrinsic property of an object, where I depends on the physical arrangement of that mass - Review example 10.10, p.312 An Unusual Baton 10.8 Energy Considerations in Rotational Motion & 10.9 Rolling Motion of a Rigid Object - The total mechanical energy is the sum of the kinetic and potential energies and obeys the principle of conservation of mechanical energy. - If an object is treated as a point particle, we can ignore the rotational motion of the particle when it is in motion - However, if the object is rotating while in motion, then we need to incorporate the rotational kinetic energy in that system - For example, for rolling disk on a flat surface both translational kinetic energy and rotational kinetic energy occur. For the rolling disk, the total kinetic energy can be expressed as K tot = 1 2 mv² Iw² where m is the mass of the object, v is the translational speed of its center of mass, I is the moment of inertia about an axis through the center of mass, and w is the angular speed.

9 - If the disk was rolling disk on a incline plane the total mechanical energy is where h is the from the floor E tot = 1 mv² + 1 Iw² + mgh 2 2 Race between a disk and hoop(=ring) Both the disk and ring was released from rest. Both objects have the same size and same mass m but the disk will always reach the bottom faster than the hoop i) Disk The total energy at top is E i=mgh and the total energy at the bottom is E tot = 1 2 mv² Iw² According to conservation of energy, the total energy is conserved throughout the motion. So mgh = 1 2 mv² Iw² For the disk, I= 1 2 mr2 and w=v / R, so mgh = 1 mv² (1 2 mr2 ) ( v R )2 Solving for v will give v= 4 gh 3 Notice, that if the object was just sliding without rotation, the speed at the bottom simply reduces to v= 2gh. ii) Hoop(or ring) For the hoop, I=mR 2, so mgh = 1 2 mv² (mr2 )( v R )2 Solving for v will give v= gh We can see the v disk = 4 gh > vhoop = gh. Hence, the translational speed of the disk is greater than the 3 hoop

10 Example) A bowling ball has a mass of 4kg and radius of 0.1m. If the is ball placed at top of an incline rolls down without slipping on the plane of angle of 22, what is the linear speed of the ball when it reaches the bottom? (you do not need all the information provided to solve the question) h=1.3m 22 - Review example 10.11, p.314 Rotating Rod Revisited Solve #55. P.329 (d) only - Review example 10.12, p.315 Energy and the Atwood Machine