Circular Motion, Pt 2: Angular Dynamics. Mr. Velazquez AP/Honors Physics

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1 Circular Motion, Pt 2: Angular Dynamics Mr. Velazquez AP/Honors Physics

2 Formulas: Angular Kinematics (θ must be in radians): s = rθ Arc Length 360 = 2π rads = 1 rev ω = θ t = v t r Angular Velocity α av = ω t = a t t Angular Acceleration T = 2π ω = 1 f Period of Rotation

3 Formulas: Angular Kinematics Linear Equation (a = constant) v = v 0 + at x = x 0 + v 0 t at2 Angular Equation (α = constant) ω = ω 0 + αt θ = θ 0 + ω 0 t αt2 v 2 = v a(x x 0 ) ω 2 = ω α(θ θ 0 ) Tangential Quantities: v t = rω a t = rα Centripetal Acceleration: a c = v t 2 r = rω2

Rolling Motion (no slipping) A perfectly round object that is rolling over a surface is experiencing static friction The part of the object in contact with the surface is always at

4 Rolling Motion (no slipping) A perfectly round object that is rolling over a surface is experiencing static friction The part of the object in contact with the surface is always at rest The linear speed v of the axle (center of the wheel) is therefore completely dependent on the angular velocity ω and the radius r of the wheel. Rolling Velocity (w/o slipping) v = rω

5 68.0 cm Example: Rolling Motion A bicycle slows down uniformly from a speed of v 0 = 8.40 m/s to rest, over a distance of 115 m. Each wheel and tire has an overall diameter of 68.0 cm. Find: a) The initial angular velocity of the wheels b) The total number of revolutions made by each wheel before coming to rest c) The angular acceleration of the wheels d) The time it took to come to a stop v 0 = 8.40 m/s v = m

6 Angular Dynamics: Torque & Moment Causing rotation obviously requires a force (Newton s 1 st Law) The direction and location of this force are highly important In the example below, a lever extends a certain length from the fulcrum (center of rotation). Several forces act on the lever. Assuming all three forces have the same magnitude, which force will cause the fastest rotation? Which force will not cause any rotation? fulcrum F C no rotation F A F B fastest rotation

7 Angular Dynamics: Torque & Moment F B will cause a greater angular acceleration for two very important reasons: Its direction is perpendicular to the lever It s applied farther from the axis of rotation (fulcrum) F C will not cause any rotation, even though it is applied farther from the fulcrum than F B This is because its direction is parallel to the lever In order to cause rotation, some component of the force must be perpendicular to the lever fulcrum F C no rotation F A F B fastest rotation

8 Angular Dynamics: Torque & Moment We can now define torque τ as the product of the perpendicular force acting on a lever and the distance away from the fulcrum this force is applied (called the lever arm) Unit: m N (meters times Newtons not the same thing as a joule, which is a N m) The greater the torque caused by a force, the greater the resulting angular acceleration (direct proportion) If the force is applied at an angle θ, we simply find the perpendicular component of that force and use it to calculate torque r ( lever arm ) F sin θ F cos θ Torque τ = rf θ F τ = rf sin θ

9 Example: Torque & Moment Two circular disks of radii r A = 30 cm and r B = 50 cm are attached to each other on an axle that passes through the center of each. Calculate the net torque on this compound disk due to the two forces shown. F A = 50 N r B r A F B = 50 N

10 Angular Dynamics: Rotational Inertia To see how torque relates to acceleration, let s examine a simple example of a particle of mass m rotating in a circle of radius r. A force F is applied to the mass in a direction tangent to the circle. The applied force will cause tangential acceleration. So we can combine the equation for tangential acceleration with the equation for Newton s Second Law: F = ma F = mrα Now, multiplying both sides by r: Torque (τ) rf = mr 2 α Moment of Inertia (I) r Torque & Moment of Inertia τ = Iα F m

11 Angular Dynamics: Rotational Inertia The moment of inertia (I) is simple for a single particle in circular motion. But a rotating rigid object (like a wheel) consists of many particles, all at different distances from the center of rotation. Therefore, the moment of inertia for a rigid object depends on the shape and density of the object and the location of the axis of rotation. The total torque on an object is related to the total moment of inertia in the following way: τ = ( mr 2 )α τ = Iα

12 Angular Dynamics: Rotational Inertia These values were computed using integral calculus; a similar table can be found in your book (Pg. 291, Table 10-1)

13 Example: Rotational Inertia Three forces are applied, as shown, to a solid circular disk of mass 3.50 kg and a radius of 25.0 cm (with an axis of rotation through the center) that is at rest. Calculate: a) The net torque acting on the disk b) The moment of inertia of the disk c) The resulting angular acceleration α F B = 55.0 N 30 F A = 60.0 N 25.0 cm 45 F B = 40.0 N

14 Rotational Kinetic Energy Any rigid, rotating object will have rotational kinetic energy (units are still joules): K R = 1 2 Iω2 Rotational Kinetic Energy If the object also undergoes translational motion as it spins (think: a tire rolling down a ramp or ball rolling across a floor), then it will have both rotational and translational kinetic energies: K total = K + K R = 1 2 mv Iω2 Total Kinetic Energy

15 Rotational Kinetic Energy Just like the work done by a force can be thought of as the change in kinetic energy, we can also define the work done by torque as the change in rotational kinetic energy: W = τ θ = 1 2 I ω2 Work Done by Torque This way, the work done by torque can be thought of as the work required to get an object spinning at a rate of ω, starting from rest.

16 Example: Rotational Kinetic Energy A spherical ball of mass M and radius R and a circular ring of the exact same mass and radius both sit at rest at the top of an incline at a vertical height of H. When they are released, which will reach the bottom of the incline first?

17 Angular Momentum When an object rotates or spins with angular speed ω, we say that it has angular momentum (L), and we calculate it exactly as expected: by replacing m and v with their angular analogues I and ω: L = Iω Angular Momentum Just as with other angular vector quantities, angular momentum can be translated into a tangential quantity that depends on the distance r from the center: L = rp Angular and Linear Momentum

18 Conservation of Angular Momentum World Record Figure Skating Spin: Angular momentum (just like linear momentum) will always be conserved. If net external torque is zero, it follows that final angular momentum will equal the initial angular momentum: L 1 = L 2 I 1 ω 1 = I 2 ω 2 Conservation of Angular Momentum

19 Exit Ticket: Rotational Dynamics A small mass (m = 1.5 kg) attached to the end of a string revolves in a circle on a frictionless tabletop. The other end of the string passes through a hole in the table. Initially, the mass revolves with a tangential speed of v 1 = 2.4 m/s in a circle of radius r 1 = 0.80 m. The string is then pulled so that the radius is reduced to r 2 = 0.48 m. Find: a) The final speed v 2 of the mass b) The initial and final angular speeds ω 1 and ω 2 c) The total kinetic energy before and after the radius changes Homework Problem Set: Rotational Dynamics Pg #44 60 (mult of 4) Pg #4 32 AND #40-60 (mult of 4)

General Definition of Torque, final. Lever Arm. General Definition of Torque 7/29/2010. Units of Chapter 10

Units of Chapter 10 Determining Moments of Inertia Rotational Kinetic Energy Rotational Plus Translational Motion; Rolling Why Does a Rolling Sphere Slow Down? General Definition of Torque, final Taking