Slide 1 / 34 Rotational Dynamics Slide 2 / 34 l Moment of Inertia To determine the moment of inertia we divide the object into tiny masses of m i a distance r i from the center. is the sum of all the tiny masses which makes up the object. This quantity is known as the moment of inertia and is denoted by I. Slide 3 / 34 Moment of Inertia Calculations: Rotating a Rod about its Center
Slide 4 / 34 Parallel Axis Theorem Figure A Figure B Figure A shows the moment of inertia of a rod rotated about its center and Figure B shows its rotation about one of its ends. If the rotation is about the center, instead of redoing the integration to find it at the end, the parallel axis theorem can be used. I cm is the moment of inertia about the rods center and d is the distance away from I cm 's rotational axis. Slide 5 / 34 1 A rod of length L when rotated about its center has a moment of inertia . What is the moment of inertia about one of its ends? A B C D Slide 6 / 34 2 A uniform stick has length L. The moment of inertia about the center of the stick is I o . A particle of mass M is attached to the stick as shown. The moment of inertia of the combined system about the center of the stick is A I o + 1/16 ML 2 I o + 1/9 ML 2 B I o + 1/4 ML 2 C L I o + 1/2 ML 2 D I o + ML 2 E L/4
Slide 7 / 34 Rigid-Body Rotation about a Moving Axis When an object moves it can have both rotational kinetic energy and translational kinetic energy. Rotational Velocity Translational Velocity For an object rolling without slipping the velocity for the center of mass is given by: Slide 8 / 34 3 A sphere rolls down the incline at height h without slipping from rest. What is the velocity of the sphere at the bottom? The moment of intertia of a sphere is . A B C D Slide 9 / 34 4 A hollow sphere initially at rest rolls down the incline at height h without slipping. What is the velocity of the hollow sphere at the bottom? The moment of intertia of a sphere is . A B C D
Slide 10 / 34 5 A solid cylinder of mass m and radius r rolls across the floor with a velocity v. Which of the following would be the best estimate of the ring's total kinetic energy as it rolls across the floor? A mv 2 1/4 mv 2 B 1/2 mv 2 C 3/4 mv 2 D 1/2 mv 2 + (mv 2 )/r E Slide 11 / 34 Work and Power in Rotational Motion We learned before that: Therefore: Slide 12 / 34 6 A conical pendulum makes 2 complete revolutions of radius r while moving with a constant angular acceleration of α. What is the work done on the bob whose moment of inertia is mr 2 ? A B C D
Slide 13 / 34 Angular Momentum The analog of linear momentum is angular momentum. Angular momentum is represented by an L, and is the product of the position vector and the momentum vector. Slide 14 / 34 Angular Momentum of a rigid body rotating about a symmetrical axis We said before that angular momentum is defined as L=mvr. If we insert v as ωr: Slide 15 / 34 Conservation of Angular Momentum When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved).
Slide 16 / 34 7 In the diagram below what is the objects angular momentum around the origin? A b B C a D Slide 17 / 34 8 An person is spinning about a vertical axis with their arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the kinetic energy and angular momentum changed? Kinetic Energy Angular Momentum A increase increase B remains constant decrease C remains constant increase increase remains constant D remains constant remains constant E Slide 18 / 34 9 An person is spinning about a vertical axis with their arms fully extended. If the arms are pulled in closer to the body, in which of the following ways are the angular velocity and moment of inertia changed? Angular Velocity Moment of Inertia increase increase A increase decrease B decrease increase C decrease decrease D
Slide 19 / 34 10 A tire of mass M and radius R rolls on a flat track without slipping. If the angular velocity of the wheel is , what is its linear momentum? M R A M 2 R B M R 2 C M 2 R 2 /2 D Zero E Slide 20 / 34 m/2 L/2 m/2 L/2 L L 2m 2m 11 The system above rotates with a speed . If the mass of the supports is negligible, what is the ratio of the angular momentum of the two upper spheres to the two lower spheres? A 16/1 C 1/16 D 1/4 B 4/1 1/1 E Slide 21 / 34 Torque r F Torque is the rotational equivalent to force. r is the length of the lever arm and it is the shortest distance from the center of rotation to the point where the force is being applied. For any type of torque we will consider the clockwise direction to be a positive torque, whereas the counter- clockwise direction is considered to be a negative torque.
Slide 22 / 34 12 What is the translational analogue of torque? A Kinetic Energy B Acceleration C Force D Mass Slide 23 / 34 13 An object of mass m is placed a distance r away from the center of a balance. If another object of mass 4m is to be placed on the balance, what distance does it have to be placed away from the center for the system to be at equilibrium? A B C D Slide 24 / 34 Calculate torque (magnitude and direction) about the pivot point of a rod of length 4m due to a force F = 10N. 90 o 40 N*m, out of page F 20 N*m, out of page 30 o F F zero F o 45 zero
Slide 25 / 34 Torque and Angular Acceleration z y Slide 26 / 34 Net Torque The net torque acting on a rotating body is equal to the derivative of the object's angular momentum. Since L can be described in terms of moment of inertia and angular velocity, the net torque can be shown as: Slide 27 / 34 A yo-yo is released with the condition that the string does not slip on the cylinder. The moment of inertia for the yo-yo is . a) Draw the free-body diagram for the yo-yo. b) Find the downward acceleration of the yo-yo. c) Find the tension in the string.
Slide 28 / 34 Answers Στ = Iα a) b) ΣF = Ma F T Στ = F T R = Iα = (1/2MR 2 )α ΣF y = F T - Mg = -Ma ΣF y = F T = Mg - Ma α = a/R F T R = (1/2MR 2 )(a/R) a F T R = 1/2MRa F T = 1/2Ma Mg F T = Mg - Ma and F T = 1/2Ma 1/2Ma = Mg - Ma 3/2Ma = Mg 3/2a = g a = 2/3g c) F T = 1/2Ma a = 2/3g F T = (1/2M)(2/3g) F T = 1/3Mg Slide 29 / 34 θ A solid sphere rolls down a ramp without slipping. The ramp has an angle of θ. The solid sphere has the moment of inertia of . a) Draw the free-body diagram for the solid sphere. b) Find the sphere's acceleration. c) Evaluate for the force of friction. Slide 30 / 34 Answers a) F N f Mg θ f = Mgsinθ - Ma and f = 2/5Ma b) ΣF = Ma Mgsinθ - Ma = 2/5Ma ΣF y = F n - Mgcosθ = Ma gsinθ = 7/5a ΣF x = Mgsinθ - f = Ma a = 5/7gsinθ f = Mgsinθ - Ma Στ = Iα Στ = fR = Iα = (2/5MR 2 )α α = a/R c) f = 2/5Ma a= 5/7gsinθ fR = (2/5MR 2 )(a/R) fR = 2/5MRa f = (2/5M) (5/7gsinθ) f = 2/5Ma f = 2/7Mgsinθ
Slide 31 / 34 14 Forces F 1 = 10.0 N and F 2 = 4.5 N are applied tangentially to a wheel with a radius of 1.5 m, as shown below. What is the net torque on the wheel due to the forces? F 1 A 6.75 N*m B 8.25 N*m 15 N*m C 1.5m 21.75 N*m D F 2 Slide 32 / 34 15 An object with moment of inertia I is originally at rest and begins to undergo a constant angular acceleration. In time t the object reaches an angular velocity of ω. What is the net torque applied to the object? A B C D Slide 33 / 34 16 I f a ladder of mass m and length L is propped up against a wall at an angle of 45 o , what is the minimum coefficient of static friction in order for the ladder not to slip? A B L C D 45 o
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