Slide 1 / 34 Rotational ynamics l Slide 2 / 34 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 alculations: Rotating a Rod about its enter Slide 4 / 34 Parallel xis Theorem igure igure igure shows the moment of inertia of a rod rotated about its center and igure 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 rod of length when rotated about its center has a moment of inertia. What is the moment of inertia about one of its ends? Slide 6 / 34 2 uniform stick has length. The moment of inertia about the center of the stick is I o. 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 I o + 1/16 M 2 I o + 1/9 M 2 I o + 1/4 M 2 I o + 1/2 M 2 E I o + M 2 /4
Slide 7 / 34 Rigid-ody Rotation about a Moving xis When an object moves it can have both rotational kinetic energy and translational kinetic energy. Slide 8 / 34 3 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. Translational Velocity Rotational Velocity or an object rolling without slipping the velocity for the center of mass is given by: Slide 9 / 34 Slide 10 / 34 4 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. 5 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? mv 2 1/4 mv 2 1/2 mv 2 3/4 mv 2 E 1/2 mv 2 + (mv 2 )/r Slide 11 / 34 Work and Power in Rotational Motion We learned before that: Slide 12 / 34 6 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? Therefore:
Slide 13 / 34 ngular Momentum The analog of linear momentum is angular momentum. ngular momentum is represented by an, and is the product of the position vector and the momentum vector. Slide 14 / 34 ngular Momentum of a rigid body rotating about a symmetrical axis We said before that angular momentum is defined as =mvr. If we insert v as ωr: Slide 15 / 34 onservation of ngular Momentum Slide 16 / 34 7 In the diagram below what is the objects angular momentum around the origin? When the net external torque acting on a system is zero, the total angular momentum of the system is constant (conserved). b a Slide 17 / 34 8 n 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? Slide 18 / 34 9 n 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? E Kinetic Energy ngular Momentum remains constant decrease remains constant remains constant remains constant remains constant ngular Velocity Moment of Inertia decrease decrease decrease decrease
Slide 19 / 34 Slide 20 / 34 10 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/2 /2 /2 m/2 M R M 2 R M R 2 M 2 R 2 /2 E Zero 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? 2m 16/1 4/1 1/16 1/4 r Slide 21 / 34 Torque E 1/1 Slide 22 / 34 12 What is the translational analogue of torque? Kinetic Energy cceleration orce Mass 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. or any type of torque we will consider the clockwise direction to be a positive torque, whereas the counterclockwise direction is considered to be a negative torque. Slide 23 / 34 Slide 24 / 34 13 n 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? alculate torque (magnitude and direction) about the pivot point of a rod of length 4m due to a force = 10N. 90 o 40 N*m, out of page 45 o 30 o 20 N*m, out of page zero zero
Slide 25 / 34 Torque and ngular cceleration z Slide 26 / 34 Net Torque The net torque acting on a rotating body is equal to the derivative of the object's angular momentum. y Since can be described in terms of moment of inertia and angular velocity, the net torque can be shown as: Slide 27 / 34 Slide 28 / 34 nswers a) T b) Σ = Ma Σy = T - Mg = -Ma Σy = T = Mg - Ma Στ = Iα Στ = TR = Iα = (1/2MR 2 )α α = a/r 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) raw the free-body diagram for the yo-yo. Mg a T = Mg - Ma and T = 1/2Ma 1/2Ma = Mg - Ma 3/2Ma = Mg 3/2a = g a = 2/3g TR = (1/2MR 2 )(a/r) TR = 1/2MRa T = 1/2Ma b) ind the downward acceleration of the yo-yo. c) T = 1/2Ma a = 2/3g c) ind the tension in the string. T = (1/2M)(2/3g) T = 1/3Mg Slide 29 / 34 Slide 30 / 34 nswers a) f N Mg θ θ solid sphere rolls down a ramp without slipping. The ramp has an angle of θ. The solid sphere has the moment of inertia of. b) Σ = Ma Σy = n - Mgcosθ = Ma Σx = Mgsinθ - f = Ma f = Mgsinθ - Ma f = Mgsinθ - Ma and f = 2/5Ma Mgsinθ - Ma = 2/5Ma gsinθ = 7/5a a = 5/7gsinθ a) raw the free-body diagram for the solid sphere. Στ = Iα Στ = fr = Iα = (2/5MR 2 )α b) ind the sphere's acceleration. c) Evaluate for the force of friction. α = a/r fr = (2/5MR 2 )(a/r) fr = 2/5MRa f = 2/5Ma c) f = 2/5Ma a= 5/7gsinθ f = (2/5M) (5/7gsinθ) f = 2/7Mgsinθ
Slide 31 / 34 14 orces 1 = 10.0 N and 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? 6.75 N*m 8.25 N*m 15 N*m 21.75 N*m 1 1.5m 2 Slide 32 / 34 15 n 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? Slide 33 / 34 Slide 34 / 34 16 If a ladder of mass m and length 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? 45 o