Rotational Motion

Rotational Motion

P r ℓ θ O Definitions • Most of our discussion will concerned with rigid bodies— objects with definite shapes that don’t change • Purely rotational motion: all points in a body move in circle

P r ℓ θ O Radians • One radian is the angle created (subtended) by an arc whose length is equal to the radius 360o = 2 π rad

Angular Velocity and Acceleration • Average angular velocity: ω = Δθ / Δt • Measured in radians per second • Average angular acceleration: α = (ω – ω0)/ Δt = Δω / Δt • Measured in radians per second squared

The Velocity of a Point • A point on a rotating wheel has the following linear velocity v = Δℓ / Δt = r (Δθ/ Δt) or v = rω

Different Points Can Have Different Velocities • Despite the fact that ω is the same for all points, points with different values of r have different velocities

Tangential Velocities Can Be Different for Points on an Object

Acceleration • Angular acceleration is related to tangential linear acceleration by: atan = Δv / Δt = r (Δω / Δt) or atan = rα • Total linear acceleration is: a = atan + aR, • Where aR is the radial or centripetal acceleration toward the center of the object’s path

Centripetal Acceleration • aR = v2/ r = (ωr)2 / r = ω2r

Frequency and Period • Frequency is the number of complete revolutions (rev) per second • One revolution corresponds to an angle of 2π radians • Therefore, 1 rev/sec = 2π rad/ sec f = ω/ 2π or ω = 2πf • The unit of frequency is the hertz (Hz) 1 Hz = 1 rev/s • The time required for one revolution is a period T T = 1/f

Centripetal Force • A force is required to keep an object moving in a circle • If the speed is constant the force is directed towards the center of the circle ∑FR = maR = mv2/r

There Is No Outward Force!

Components of Circular Motion

Force On a Revolving Ball (Horizontal) • Estimate the force a person must exert on a string attached to a 0.15 kg ball to make the ball revolve in a horizontal circle of radius 0.6 m. The ball makes 2 revolutions per second • F = mv2/r = (0.15 kg)(7.54 m/s)2/(0.6 m) ≈ 14 N • This solution ignores the fact that the ball cannot be perfectly horizontal because it has weight due to the force of gravity

Force On a Revolving Ball (Vertical) • A 0.15 kg ball on the end of a 1.1 m string is swung in a vertical circle. Determine the minimum speed the ball must have at the top of its arc so that it continues moving in a circle. Calculate the tension in the string at the bottom of the arc if the ball is moving at twice the minimum speed.

Solution • At the top of the arc there are two forces on the ball: mg and the tension FTA ∑FR = maR FTA + mg = mvA2/r • The minimum speed occurs when FTA = 0 mg = mvA2/r vA= √(gr) = 3.28 m/s

Solution (cont’d) • At the bottom of the circle the cord exerts its tension force FTB upward, but the force of gravity mg is downward ∑FR = maR FTB – mg = mvB2/r = mvB2/r + mg • For v = 6.56 m/s (2x minimum) F= 7.34 N

Ferris Wheel • A rider on a Ferris wheel moves in a vertical circle of radius r at constant speed v • Is the normal force the seat exerts on the rider at the top of the circle less than, more than, or the same as the force exerted at the bottom of the arc?

Solution • This is exactly like the vertical string problem with FN replacing tension. Therefore, the force at the top is less than the force at the bottom

Centripetal Forces--Uniform Circular Motion

Roller Coaster Physics

Demonstrations

A Right Hand Turn

Forces on Cars In Turns Centripetal acceleration is horizontal, not parallel to road surface Flat Road Banked Turn

Banked vs Flat Turns

Circular Motion as the Limit of Straight Line Motion

Ball on a String • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c8_whirligig.html

Gravitron • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c7_rotor.html

Objects on a Turntable • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c7_turntable.html

Vertical Circular Motion • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c8_vertical.html

Angular ω = ω0 + ½αt θ = ω0t + ½αt2 ω2 = ω02 + 2αθ ω = (ω + ω0)/2 Linear v = v0 + at x = v0t + ½ at2 v2 = v02 + 2ax v = (v + v0)/2 Kinematic Equations

Example • A bicycle slows down from 8.4 m/s to rest over a distance of 115 m. Each wheel has a diameter of 68.0 cm. • Determine the angular velocity of the wheels at the initial moment; the total number of revolutions each wheel makes in coming to rest, and the time it took to stop.

Solution • At the initial instant, points on the rim of the wheel are moving at 8.4 m/s. • The initial angular velocity is: ω0 = v0/r = (8.4 m/s)/ (0.34 m) = 24.7 rad/s • 115m of ground passes under the bike as it stops. Each revolution of a wheel corresponds to a distance of 2πr so: 115 m/ 2πr = 115 m/(2πr)(0.34 m) = 53.8 rev

Solution • Angular acceleration of the wheel can be obtained from: ω2 = ω02 + 2αθ α = (ω2 - ω02)/ 2θ = 0 – (24.7 rad/s)2 2(2π)(53.8 rev) = -0.902 rad/s2 From ω = ω0 + ½αt we get that: t = (ω - ω0)/α = (0 – 24.7 rad/s)/ -0.902 rad/s2 = 27.4 s

Rolling • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c15_rolling.html

Rotational Dynamics • Up to this point we have been studying rotational kinematics— how things move • Now we will go on to rotational dynamics—why things move

Ferris Wheel Kinematics • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c13_consta_ex.html

Torque • Causing an object to rotate around its axis requires a force • The effect of a force is greater if it is placed further from the axis of rotation • The angular acceleration of an object is directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts • This distance is called a lever arm • The product of the force times the lever arm is called torque (α proportional to τ)

Only the Perpendicular Component of Force Contributes to Torque Torque = r┴F = rF┴

Example • The biceps muscle exerts a vertical force on the lower arm. Calculate the torque about the axis of rotation through the elbow joint assuming the muscle is attached 5.0 cm from the elbow

Do Now (10/1/13): • The biceps muscle exerts a vertical force on the lower arm. Calculate the torque about the axis of rotation through the elbow joint assuming the muscle is attached 5.0 cm from the elbow

Solution F = 700 N and r┴ = 0.05m so τ = r┴F = (0.05 m)(700 N) = 35 m-N

Solution r┴ = (0.05 m)(sin 60o) Therefore τ = (0.05 m)(sin 60o)(700N) = (0.05 m)(0.866)(700N) = 30 m-N

Torque on a Compound Wheel • Two thin cylindrical wheels of radii r1 = 30 cm and r2 = 50 cm are attached to each other as shown. Calculate the net torque on this wheel due to the two forces shown, each of magnitude 50 N

Solution • The two forces create torques in different directions • We can consider one to be positive an done to be negative • Because F2 is not perpendicular to the axis of rotation we must only use the component of the force that is perpendicular τ = r1F1 – r2F2sin60o = (0.3m)(50N) – (0.5m)(50 N)(0.866)= -6.7 m-N

Balancing Torques Force = F Force = 2F C B A D X X A force of magnitude F is applied at a distance X from the center of a seesaw. Another force of magnitude 2F is also applied to the seesaw at a distance X on the other side of the fulcrum. At what location(s) and in what directions can a third force of magnitude F be applied so that the seesaw is balanced?

Solution Force = F Force = F Force = 2F C B A D Force = -F X X The seesaw can be balanced if a force of magnitude F is applied in the positive direction at A or in the negative direction at D

Torques • http://webphysics.davidson.edu/physlet_resources/bu_semester1/c14_equilibrium.html

Rotational Analogs of Mass and Momentum • The angular acceleration of an object is directly proportional to the perpendicular distance from the axis of rotation to the line along which the force acts • α is proportional to Στ

Rotational Motion