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Angular Position

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Angular Position ANGULAR POSITION • To describe rotational motion, we define angular quantities that are analogous to linear quantities • Consider a bicycle wheel that is free to rotate about its axle • The axle is the axis of rotation for the wheel • If there is a small spot of red paint on the tire, we can use this reference to describe its rotational motion • The angular position of the spot is the angle θ, that a line from the axle to the spot makes with a reference line • SI unit is the radian (rad) • θ > 0 – anticlockwise rotation: θ < 0 – clockwise rotation • A radian is the angle for which the arc length, s, on a circle of radius r is equal to the radius of the circle • The arc length s for an arbitrary angle θ measured in radians is s = r θ • 1 revolution is 360°= 2 π rad • 1 rad = 360°/2 π = 57.3° 4. Rotational Kinematics 1 and Dynamics ANGULAR VELOCITY • As the bicycle wheel rotates, the angular position of the spot changes ∆ • Angular displacement is θ = θf – θi ∆ ∆t • Average angular velocity is ωav = θ/ (rad/s) v ∆x ∆t • Analogous average linear velocity av = / • Instantaneous angular velocity is the limit of ωav as the time interval ∆t reaches zero • ω > 0 – anticlockwise rotation: ω < 0 clockwise rotation • The time to complete one revolution is known as the period , T • T = 2 π/ω seconds 4. Rotational Kinematics 2 and Dynamics ANGULAR ACCELERATION • If the angular velocity of the rotating bicycle wheel increases or decreases with time, the wheel experiences an angular acceleration , α • The average angular acceleration is the change in angular velocity in a given time interval ∆ ∆t 2 • αav = ω/ rad/s • The instantaneous angular acceleration is the limit of ∆t αav as the time interval approaches zero • The sign of angular acceleration is determined by whether the change in angular velocity is positive or negative • If ω is becoming more positive (ωf > ωi), α is positive • If ω is becoming more negative (ωf < ωi), α is negative • If ω and α have the same sign, speed of rotation increasing • If ω and α have opposite signs, speed of rotation decreasing 4. Rotational Kinematics 3 and Dynamics ROTATIONAL KINEMATICS • Rotational kinematics describes rotational motion • Consider the pulley shown below, which has a string wrapped around its circumference with a mass attached to its free end • When the mass is released, the pulley begins to rotate – slowly at first, then faster and faster • The pulley thus accelerates with constant angular acceleration: α = ∆ω/∆t • If the pulley starts with initial angular velocity ω0 at time t = 0, and at the later time t the angular velocity is ω ∆ ∆t t t t then α = ω/ = ( ω – ω0)/( – 0) = (ω – ω0)/ • Thus the angular velocity ω varies with time as follows: t ω = ω0 + α • Example: If the angular velocity of the pulley is -8.4rad/s at a given time, and its angular acceleration is -2.8rad/s 2, what is the angular velocity of the pulley 1.5s later? 4. Rotational Kinematics 4 and Dynamics LINEAR AND ANGULAR ANALOGIES 4. Rotational Kinematics 5 and Dynamics ROTATIONAL KINEMATICS: EXAMPLE (1) • To throw a curve ball, a baseball pitcher gives the ball an initial speed of 36.0 rad/s. When the catcher gloves the ball 0.595s later, its angular speed has decreased (due to air resistance) to 34.2 rad/s. What is the ball’s angular acceleration, assuming it to be constant? How many revolutions does the ball make before being caught? 4. Rotational Kinematics 6 and Dynamics ROTATIONAL KINEMATICS: EXAMPLE (2) • On a TV game show, contestants spin a wheel when it is their turn. One contestant gives the wheel an initial angular speed of 3.4 rad/s. It then rotates through1 ¼ revolutions and comes to rest on the BANKRUPT space. Find the angular acceleration of the wheel, assuming it to be constant. How long does it take for the wheel to come to a rest? 4. Rotational Kinematics 7 and Dynamics TANGENTIAL SPEED OF A ROTATING OBJECT • Consider somebody riding a merry-go-round, which completes one circuit every T = 7.5s • Thus ω = 2 π/T = 0.838 rad/s • The path followed is circular, with the centre of the circle at the axis of rotation • The rider is moving in a direction that is tangential to the circular path • The tangential speed is the speed at a tangent to the circular path, and is found by dividing the circumference T: v r T by t = 2 π / m/s T v r • Because 2 π/ = ω we have: t = ω m/s • Example: Find the angular speed a CD must have to give a linear speed of 1.25m/s when the laser beam shines on the disk 2.50cm and 6.00cm from its centre 4. Rotational Kinematics 8 and Dynamics CENTRIPETAL ACCELERATION OF A ROTATING OBJECT • When an object moves in a circular path, it experiences a a centripetal acceleration, cp , which is always directed toward the axis of rotation a v2 r • cp = / v v r a r 2 r = r 2 2 • However = t = ω, so cp = ( ω) / ω m/s • Rotating devices known as centrifuges can produce centripetal accelerations many times greater than gravity, such as those used to train astronauts, or microhematocrit centrifuges used to separate blood cells from plasma • Example: In a microhematocrit centrifuge, small samples of blood are placed in capillary tubes. These tubes are rotated at 11,500rpm, with the bottom of the tubes 9.07cm from the axis of rotation. Find the linear speed of the bottom of the tubes. What is the centripetal acceleration at the bottom of the tubes? 4. Rotational Kinematics 9 and Dynamics TANGENTIAL AND CENTRIPETAL ACCELERATION • When the angular speed of an object in a circular path changes, so does its tangential speed • When tangential speed changes, a tangential a acceleration is experienced t • If ω changes by the amount ∆ω, with r remaining constant, the corresponding change in tangential speed ∆v r∆ is t = ω • If ∆ω occurs in time interval ∆t, then the tangential a ∆v ∆t r∆ ∆t acceleration is t = t/ = ω/ • Since ∆ω/∆t is the angular acceleration α, then the tangential acceleration of a rotating object is given a r 2 by t = α m/s a • Recall that t is due to a changing tangential speed, a and that cp is caused by a changing direction of motion a (even if t remains constant) • In cases where both tangential and centripetal accelerations are present, the total sum is the vector sumr of ther two a a •t and cp are at right angles, hence the magnitude of a √ a 2 a 2 the total acceleration is = ( t + cp ) -1 a a • The direction is given by φ = tan ( cp / t) 4. Rotational Kinematics 10 and Dynamics TANGENTIAL AND CENTRIPETAL ACCELERATION: EXAMPLE • Suppose the centrifuge above is starting up with a constant angular acceleration of 95.0 rad/s 2. What is the magnitude of the centripetal, tangential and total accelerations of the bottom of a tube when the angular speed is 8.00 rad/s? What angle does the total acceleration make with the direction of motion? 4. Rotational Kinematics 11 and Dynamics TORQUE: WHEN FORCE APPLIED IS TANGENTIAL • Trying to loosen a nut by rotating a wrench anticlockwise is easier when you apply the force as far away from the nut as possible • Likewise to open a revolving door is easier when you push further from the axis of rotation • The tendency for a force to cause a rotation increases with the distance r from the axis of rotation to the force • Torque is a quantity that takes into account both the magnitude of the force and the distance from the axis of rotation, r • Torque: τ = rF Nm (Newton – metre) • This equation is only valid when the applied force is tangential to a circle of radius r centred on the axis of rotation 4. Rotational Kinematics 12 and Dynamics TORQUE: WHEN FORCE APPLIED IS NOT TANGENTIAL • Consider pulling on a merry-go-round in a direction that is radial (along a line that extends through the axis of rotation) • Such a force has no tendency to cause a rotation, and thus the axle simply exerts an equal and opposite force, and thus the merry-go-round remains at rest • A radial force produces zero torque • If the force appliedr is at an angle θ to the radial line, the vector force F needs to be resolved into radial and tangential components • Radial component magnitude: Fcos θ • Tangential component magnitude: Fsin θ • Only tangential component causes rotation, thus Fcos θ = 0 • General definition of torque: τ = rF sin θ Nm • τ > 0 – anticlockwise angular acceleration • τ < 0 – clockwise angular acceleration 4. Rotational Kinematics 13 and Dynamics TORQUE: EXAMPLE • Two helmsmen, in disagreement about which way to turn a ship, exert different forces on the ship’s wheel. The wheel has a radius of 0.74m, and the two forces F F have the magnitudesr 1 = 72N and 2 = 58N. Findr the F F torque caused by 1 and the torque caused by 2 . In which direction does the wheel turn as a result of these two forces. 4. Rotational Kinematics 14 and Dynamics TORQUE AND ANGULAR ACCELERATION • A single torque, τ, acting on an object causes the object to have an angular acceleration α • Consider a small object of mass m connected to an axis of rotation by a light rod of length r • If a tangential force of magnitude F is applied to the mass, it will move with an acceleration according to Newton’s 2 nd law, a = F/m • Linear and angular accelerations related by α = a/r • Combining: α = a/r = F/mr • Multiplying by r/r gives α = rF /mr 2 • Since torque τ = rF , we define a new quantity called the moment of inertia : I = mr 2 • Thus α = τ/I or τ = Iα • In a system with more than one torque, we take the net Σ I sum of all the torques acting: τnet = τ = α • Above is Newton’s 2 nd law for rotational motion 4.
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