Physics 201. Professor P. Q. Hung. 311B, Physics Building. Physics 201 p. 1/1

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1 Physics 201 p. 1/1 Physics 201 Professor P. Q. Hung 311B, Physics Building

2 Physics 201 p. 2/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia All elements inside the rigid body have the same angular velocity ω. Also v i = r i ω

3 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia All elements inside the rigid body have the same angular velocity ω. Also v i = r i ω An element m i at a distance r i from the axis of rotation has a kinetic energy K i = 1 2 m iv 2 i = (1 2 m ir 2 i )ω2 Physics 201 p. 2/1

4 Physics 201 p. 3/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia Total kinetic energy of the rigid body: K tot = ( 1 2 m ir 2 i )ω2 = 1 2 ( m i r 2 i )ω2

5 Physics 201 p. 3/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia Total kinetic energy of the rigid body: K tot = ( 1 2 m ir 2 i )ω2 = 1 2 ( m i r 2 i )ω2 Moment of Inertia: I = m i r 2 i K tot = 1 2 Iω2

6 Physics 201 p. 3/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia Total kinetic energy of the rigid body: K tot = ( 1 2 m ir 2 i )ω2 = 1 2 ( m i r 2 i )ω2 Moment of Inertia: I = m i r 2 i K tot = 1 2 Iω2 Notice the correspondances: I m; ω v

7 Physics 201 p. 4/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia Some examples of moments of inertia Dumbell rotating about its center: I = mr 2 + mr 2 = 2mr 2

8 Physics 201 p. 5/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia Hoop: I = m i r 2 i = ( m i )R 2 = MR 2

9 Physics 201 p. 6/1 Rotational Kinematics and Energy Rotational Kinetic Energy, Moment of Inertia

10 Physics 201 p. 7/1 Rotational Kinematics and Energy Conservation of Energy Conservation of Energy: U + K = constant

11 Physics 201 p. 7/1 Rotational Kinematics and Energy Conservation of Energy Conservation of Energy: U + K = constant Rolling without slipping: K = 1 2 mv Iω2 = 1 2 mv2 (1 + I mr 2 )

12 Physics 201 p. 7/1 Rotational Kinematics and Energy Conservation of Energy Conservation of Energy: U + K = constant Rolling without slipping: K = 1 2 mv Iω2 = 1 2 mv2 (1 + I mr 2 ) For example for a rolling sphere: I = 2 5 mr2 K = 1 2 mv2 ( ) = 7 10 mv2.

13 Physics 201 p. 7/1 Rotational Kinematics and Energy Conservation of Energy Conservation of Energy: U + K = constant Rolling without slipping: K = 1 2 mv Iω2 = 1 2 mv2 (1 + I mr 2 ) For example for a rolling sphere: I = 2 5 mr2 K = 1 2 mv2 ( ) = 7 10 mv2. A rolling hoop: I = mr 2 K = mv 2

14 Physics 201 p. 8/1 Rotational Kinematics and Energy Conservation of Energy A sphere and a hoop (of same radius) rolling down an inclined plane of height h. A block sliding down the same plane: U i + K i = mgh + 0; U f + K f = mv2 mgh = 1 2 mv2 v = 2gh

15 Physics 201 p. 8/1 Rotational Kinematics and Energy Conservation of Energy A sphere and a hoop (of same radius) rolling down an inclined plane of height h. A block sliding down the same plane: U i + K i = mgh + 0; U f + K f = mv2 mgh = 1 2 mv2 v = 2gh Rolling sphere (without slipping): mgh = 7 10 mv2 10 v = 7 gh < 2gh

16 Physics 201 p. 8/1 Rotational Kinematics and Energy Conservation of Energy A sphere and a hoop (of same radius) rolling down an inclined plane of height h. A block sliding down the same plane: U i + K i = mgh + 0; U f + K f = mv2 mgh = 1 2 mv2 v = 2gh Rolling sphere (without slipping): mgh = 7 10 mv2 10 v = 7 gh < 2gh A hoop: mgh = mv 2 v = gh

17 Physics 201 p. 9/1 Rotational Kinematics and Energy Conservation of Energy Which one is fastest? mgh = 1 2 mv Iω2

18 Physics 201 p. 9/1 Rotational Kinematics and Energy Conservation of Energy Which one is fastest? mgh = 1 2 mv Iω2 Hoop: K rotation = 1 2 I hoopω 2 ; Sphere (of same radius): K rotation = 1 2 I sphereω 2

19 Physics 201 p. 9/1 Rotational Kinematics and Energy Conservation of Energy Which one is fastest? mgh = 1 2 mv Iω2 Hoop: K rotation = 1 2 I hoopω 2 ; Sphere (of same radius): K rotation = 1 2 I sphereω 2 K rotation,hoop > K rotation,sphere Less energy from mgh converted into the translation K.E. for the hoop It arrives at the bottom last.

20 Physics 201 p. 10/1 Rotational Dynamics and Static Equ Torque You want to rotate a wheel from rest, to loosen a nut, etc.. What do you do?

21 Physics 201 p. 11/1 Rotational Dynamics and Static Equ Torque Look at Fig For the most efficient way of doing that, we would apply the force in the direction which is perpendicular to the wheel poke or to the wrench. For other angles, the effect of the applied force will be less. In addition, it turns out that it is more efficient if it is applied as far as possible from the axis of rotation. This tells us that we should look for a quantity that can exhibit such a behaviour.

22 Physics 201 p. 12/1 Rotational Dynamics and Static Equ Torque Furthermore, as long as we keep applying the force, the object (wheel, wrench,..) turns faster and faster. That means it has an angular acceleration α. For linear dynamics, a net force gives rise to an linear acceleration. What gives rise to α? The Torque: τ. It is also a vector!

23 Physics 201 p. 13/1 Rotational Dynamics and Static Equ Torque Torque: τ = rf sin θ; θ: angle of the force relative to the radial line.

24 Physics 201 p. 13/1 Rotational Dynamics and Static Equ Torque Torque: τ = rf sin θ; θ: angle of the force relative to the radial line. The force F can be decomposed into a component F cos θ, along the radial line and a component F sin θ perpendicular to the radial line and tangential to the trajectory. According to the above statement, only the latter component contributes to the torque.

25 Physics 201 p. 14/1 Rotational Dynamics and Static Equ Torque

26 Physics 201 p. 15/1 Rotational Dynamics and Static Equ Torque Torque as a vector: τ = r F : Vector product.

27 Physics 201 p. 15/1 Rotational Dynamics and Static Equ Torque Torque as a vector: τ = r F : Vector product. C = A B. C perpendicular to both A and B. C = AB sin θ.

28 Physics 201 p. 16/1 Rotational Dynamics and Static Equ Torque and Angular acceleration Take an object of mass m and attach it to one end of a massless rod of length r. Apply a force F on the object. As we have seen, only the component F t = ma t tangential to the trajectory contributes. The torque on the particle is τ = rf t = rma t. But a t = rα

29 Physics 201 p. 16/1 Rotational Dynamics and Static Equ Torque and Angular acceleration Take an object of mass m and attach it to one end of a massless rod of length r. Apply a force F on the object. As we have seen, only the component F t = ma t tangential to the trajectory contributes. The torque on the particle is τ = rf t = rma t. But a t = rα τ = mr 2 α = Iα or τ = I α

30 Physics 201 p. 17/1 Rotational Dynamics and Static Equ Torque and Angular acceleration: Example A rigid massless rod of length L = 0.83 m has two objects of mass m = 0.25 kg each attached to each end of the rod as shown in class. The rod rotates with the axis of rotation going through the center of the rod. The initial angular speed is ω 0 = 2.1 rad/s. A force F = 9.6 N is applied to one of the two objects and tangential to the circular trajectory, and for t = 2 s. Find the final ω

31 Physics 201 p. 18/1 Rotational Dynamics and Static Equ Torque and Angular acceleration: Example Use ω = ω 0 + αt. What s α?

32 Physics 201 p. 18/1 Rotational Dynamics and Static Equ Torque and Angular acceleration: Example Use ω = ω 0 + αt. What s α? α = τ/i. Find τ and I.

33 Physics 201 p. 18/1 Rotational Dynamics and Static Equ Torque and Angular acceleration: Example Use ω = ω 0 + αt. What s α? α = τ/i. Find τ and I. τ = rf = (L/2)F and I = m( L 2 )2 + m( L 2 )2 = 1 2 ml2.

34 Physics 201 p. 18/1 Rotational Dynamics and Static Equ Torque and Angular acceleration: Example Use ω = ω 0 + αt. What s α? α = τ/i. Find τ and I. τ = rf = (L/2)F and I = m( L 2 )2 + m( L 2 )2 = 1 2 ml2. α = F ml.

35 Physics 201 p. 18/1 Rotational Dynamics and Static Equ Torque and Angular acceleration: Example Use ω = ω 0 + αt. What s α? α = τ/i. Find τ and I. τ = rf = (L/2)F and I = m( L 2 )2 + m( L 2 )2 = 1 2 ml2. α = F ml. ω = ω 0 + F ml t = 94.6rad/s

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