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**Chapter 7 Circular Motion and Gravitation**

Our discussion of motion up until now has been concerned entirely with translation, in which an object moves along a straight or curved line. In this chapter, we will introduce the concept of circular motion, in which an object follows a path with a constant radius of curvature. After introducing the rotational variables (angular displacement, angular velocity, etc.), we will develop equations to describe the evolution of these variables with time. Finally, we will discuss a rotational version of Newtonβs 2nd law, and its application to gravitation. PC141 Intersession 2013

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7.1 Angular Measure In order to quantify circular motion, we need to introduce new units of angular measure. Since a circle lies in a plane, we need to work in two dimensions. As we know, any point (P) in 2D can be described by x- and y-coordinates (these are known as Cartesian coordinates). However, circular motion is best described using polar coordinates, r and π. r is the distance of P from the origin, while π is its angle, measured counterclockwise from the positive x-axis. PC141 Intersession 2013

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7.1 Angular Measure From the figure (and basic trigonometryβ¦the same math that we used when describing vector components in chapter 3), we see that π₯=π cos π π¦=π sin π π= π₯ 2 + π¦ 2 π= tan β1 π¦ π₯ (as usual, we need to make sure that the correct solution of the arctangent function is used) PC141 Intersession 2013

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7.1 Angular Measure If an object is moving in a circular path, then r is constant, and only π½ changes with time. For many problems, we are therefore only concerned with the latter variable. In chapter 2, we defined displacement (which, in the present context, should really be termed linear displacement): βπ₯=π₯β π₯ 0 . Analagously, we can define angular displacement: βπ=πβ π 0 (often, we let the initial angle π 0 be zero, in which case we simply have βπ=π). PC141 Intersession 2013

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7.1 Angular Measure We are all familiar with the most common unit for measuring angles β the degree (there are 360Β° in one full revolution). However, a much more useful unit is the radian (rad). This unit arises when one considers the relation between angular displacement and the arc length (s) over which a particle lying a distance r from the origin travels. By definition, the angle in radians is the ratio of s to r. That is, π =ππ As the ratio of two distances, the radian is actually dimensionless (the text also calls it unitless, but this is purely an issue of terminology. Iβd argue that βradianβ is the unit). PC141 Intersession 2013

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7.1 Angular Measure Converting between radians and degrees is quite simple. We know that the circumference of a circle is 2ππ. Therefore, in one full revolution, there are 360Β° and π= π π = 2ππ π =2π radians. For angles that are simple multiples or fractions of Ο, it is common to avoid writing out the digits. For example, you would say that 1.5 revolutions is β3π radiansβ rather than β9.42 radiansβ, although the latter is correct. PC141 Intersession 2013

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**Problem #1: Car Tire Revolutions**

WBL Ex 7.9 A car with a 65-cm-diameter wheel travels 3.0 km. How many revolutions does the wheel make in this distance? Solution: In class PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

Angular speed and velocity are defined analogously to their translational counterparts from chapter 2. Average angular speed is calculated as π = βπ βπ‘ = πβ π 0 π‘β π‘ 0 (as usual, the overbar indicates an average quantity. π is the Greek lower-case letter βomegaβ). The SI unit for π (and all other angular speeds and velocities) is the radian per second. If we consider only an infinitesimally short time interval, we have the instantaneous angular speed, π. If π is constant, then π =π. Then, if we set π 0 and π‘ 0 to be zero, we simply have π= π π‘ or π=ππ‘ PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

There is another common unit for angular speed, the revolution per minute (rpm). To convert between the two: 1 revolution min 1 min 60 s 2Ο rad 1 revolution = π 30 rad/sβ0.105 rad/s The average and instantaneous angular velocity are vectors; they imply both an angular speed and an angular direction. Recall from chapter 2 that in one dimension, velocities can only be positive or negative (depending on whether the object is moving toward increasing or decreasing values of x). Similarly, for angular motion, there are only two possible βdirectionsββ¦ toward increasing or decreasing values of π. PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

Actually defining the direction of angular motion is a bit tricky. Picture a few ants sitting on a turntable, which lies in the xy-plane. At any moment in time, each ant is moving in a different direction. However, all of them are rotating either clockwise or counterclockwise (Iβm pretty sure that turntables rotate clockwise, but I havenβt owned a turntable since 1989 and Iβm not willing to hit up a club just for the sake of making these notes). However, we canβt really say that the motion is βclockwiseβ, because this depends on which side of the turntable weβre looking at. If we were to somehow get below the turntable and then look back up at it, we would claim that it rotates counterclockwise. PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

To remove this ambiguity, we apply the right-hand rule, which is familiar to anyone who has learned about the cross-product of vectors. Curl the fingers of your right hand in the direction of the rotation. π points in the direction of your right thumb. This is how we define the βdirectionβ of angular velocity β even though no portion of the rotating object actually moves in this direction at any time! PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

Relation between Tangential and Angular Speeds A particle moving in a circle has an instantaneous velocity tangential to its circular path. The direction of this velocity is constantly changing, and is generally irrelevant. However, the tangential speed ( π£ π‘ ) is quite important for various reasons. For now, we will assume that the angular velocity is constant. Combining equations from slides 5 and 8, we find that the arc length is π =ππ=π(ππ‘) Then, because distance is speed multiplied by time (when speed is constant), we can also write the arc length as π = π£ π‘ π‘. Combining these equations produces π£ π‘ =ππ PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

This equation tells us that, while all particles in a rotating object have the same angular velocity, their tangential speeds depend on their distance from the origin. PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

Period and Frequency Circular motion with constant angular speed results in a periodic repeating of the same motion β all particles in a rotating object simply travel along the same circular path over and over again. The time required to complete one rotation (or cycle) is called the period (T) of the motion. For example, the period of the Earthβs rotation about its axis is 24 hoursβ . A related parameter is the frequency (f) of the circular motion. This is the number of revolutions in a given amount of time. the SI unit for frequency is s-1 (the βinverse secondβ), which is also called the Hertz (Hz). An object that completes 10 revolutions in 2 seconds has T = 0.2 s and f = 5 Hz. β more or less (donβt ask) PC141 Intersession 2013

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**7.2 Angular Speed and Velocity**

Based on their definitions, itβs clear that the two quantities defined on the previous page are simply the inverse of each other: π= 1 π The tangential speed is related to the period by π£ π‘ =2ππ/π. Since π£ π‘ =ππ, we can also write that π= 2π π =2ππ PC141 Intersession 2013

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**Problem #2: Merry-Go-Round**

WBL Ex 7.27 A little boy jumps onto a small merry-go-round (radius of 2.00 m) in a park and rotates for 2.30 s through an arc length distance of 2.55 m before coming to rest. If he landed (and stayed) at a distance of 1.75 m from the central axis of rotation of the merry-go-round, what was his average angular speed and average tangential speed? Solution: In class PC141 Intersession 2013

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**Problem #3: Measuring the Speed of Light**

The speed of light is measured as shown in the figure. A slotted wheel is rotated while light passes through the slots. The light travels to a mirror a distance L away and then returns to the wheel just in time to pass through the next slot. For this particular wheel, the radius is 5.0 cm and there are 500 slots around the edge. The mirror is L = 500 m from the wheel. The speed of light is measured at 3 x 108 m/s. What is the constant angular speed of the wheel, and what is the linear speed of a point on the edge of the wheel? Solution: In class PC141 Intersession 2013

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**7.3 Uniform Circular Motion and Centripetal Acceleration**

Uniform circular motion occurs when an object moves at a constant speed in a circular path. Note that the object does not need to complete any full revolutions. Although the objectβs speed is constant, its velocity is not, since it is constantly changing direction. Therefore, there must be an acceleration associated with this motion. PC141 Intersession 2013

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**7.3 Uniform Circular Motion and Centripetal Acceleration**

What is the direction of this acceleration? Itβs definitely not in the direction of the instantaneous velocity (tangent to the circular path), since this would imply that the object is speeding up β the circular motion wouldnβt be uniform. In fact, as the figure indicates, β v , and therefore the acceleration, are directed radially inward, toward the center of the circle. This is termed centripetal acceleration, a π (the term βcentripetalβ means βcenter-seekingβ). PC141 Intersession 2013

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**7.3 Uniform Circular Motion and Centripetal Acceleration**

The magnitude of centripetal acceleration, ac, is derived on p. 231 of the text. It can be expressed as a function of tangential speed and the circleβs radius: π π = π£ 2 π or as a function of angular speed and radius: π π =π π 2 (the text drops the subscript on tangential speed at this point). These equations beg the question: is centripetal acceleration directly proportional, or inversely proportional to the radius? Weβll discuss it in class. PC141 Intersession 2013

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**7.3 Uniform Circular Motion and Centripetal Acceleration**

Centripetal Force Newtonβs 2nd law tells us that where thereβs an acceleration, there must be a force. Since F = ma, we can define the centripetal force: πΉ π =π π π = π π£ 2 π =ππ π 2 (the direction of Fc is the direction of ac radially inward). Since the centripetal force is perpendicular to the instantaneous velocity (which is in the tangential direction), it can do no work on the object. Therefore, the objectβs KE does not change, and the objectβs speed remains constant. Which is a requirement for uniform circular motion. Everything checks out! PC141 Intersession 2013

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**7.3 Uniform Circular Motion and Centripetal Acceleration**

Centripetal Force contβ It is important to note that the centripetal force is not a result of the uniform circular motion, it is the cause. If an object is moving along a circular path, there must be a reason. For example, a satellite orbiting the Earth maintains a circular path because of the gravitational force (which points inward, toward the center of the Earth). A rock that is swung at the end of a rope is subjected to a (radially inward) tension force along the rope. A car that travels around a curve experiences (radially inward) static frictional forces between the tires and the road. Should fs,max be reached, the car will skid in a straight, tangential line. If the rope breaks, the rock is flung along a straight, tangential line. If the Earth disappears, weβre all screwed the satellite will travel in a straight line. PC141 Intersession 2013

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**Problem #4: Uniform Circular Motion**

WBL LP 7.9 In uniform circular motion, there is aβ¦ A β¦constant velocity B β¦constant angular velocity C β¦zero acceleration D β¦nonzero tangential acceleration PC141 Intersession 2013

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**Problem #5: Stunt Flying**

WBL Ex 7.33 An airplane pilot is going to demonstrate flying in a tight vertical circle. To ensure that she doesnβt black out at the bottom of the circle, the acceleration must not exceed 4.0g. If the speed of the plane is 50 m/s at the bottom of the circle, what is the minimum circle radius so that the 4.0g limit is not exceeded? Solution: In class PC141 Intersession 2013

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**Problem #6: Loop-the-Loop**

WBL Ex 7.41 A block of mass m slides down an inclined plane into a loop-the-loop of radius r. Neglecting friction, what is the minimum speed the block must have at the highest point of the loop in order to stay in the loop? At what vertical height on the inclined plane (in terms of r) must the block be released if it is to have the required minimum speed at the top of the loop? Solution: In class PC141 Intersession 2013

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**Problem #7: Swinging a Rock**

A string 1.00 m in length has a breaking strength of 50.0 N. A rock with mass 0.10 kg is tied to one end. The other end is held tightly, and the rock is swung in circular motion along a horizontal plane. What is the maximum angular velocity, in rpm, that the rock can be swung without breaking the string? Solution: In class PC141 Intersession 2013

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