Chapter 6: Uniform Circular Motion PDF

Chapter 6 : Uniform circular motion Introduction Uniform circular motion, motion in a circular path at constant speed. Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Rotation ang...

Chapter 6 : Uniform circular motion Introduction Uniform circular motion, motion in a circular path at constant speed. Pure rotational motion occurs when points in an object move in circular paths centered on one point. Pure translational motion is motion with no rotation. Rotation angle and Angular Velocity In Kinematics, we studied motion along a straight line and introduced such concepts as displacement, velocity, and acceleration. In this chapter, uniform circular motion by defining two angular quantities needed to describe rotational motion. Rotation Angle When objects rotate about some axis—for example, when the CD (compact disc) rotates about its center—each point in the object follows a circular arc. Consider a line from the center of the CD to its edge. Each pit used to record sound along this line moves through the same angle in the same amount of time. The rotation angle is the amount of rotation and is analogous to linear distance. We define the rotation angle to be the ratio of the arc length to the radius of curvature: ∆𝑆 ∆𝜃 = 𝑟  The arc length ∆𝑆 is the distance traveled along a circular path and r is the radius of curvature of the circular path. For one complete revolution, the arc length is the circumference of a circle of radius. The circumference of a circle is 2𝜋𝑟. Thus, for one complete revolution the rotation angle is 2𝜋𝑟 ∆𝜃 = = 2𝜋 𝑟 ∆𝜃 to be radians (rad), defined so that 2𝜋 rad = 1 revolution. For a circle there is 360o , that means if a CD completes one revolution, 2𝜋 rad = 360o 1 rad = 360o /2 𝜋 ≈ 57.3o Angular Velocity To describe how fast the object rotating we define angular velocity. Angular velocity 𝜔 is defined as the rate of change of an angle. ∆𝜃 𝜔= ∆𝑡 The greater the rotation angle in a given amount of time, the greater the angular velocity. The units for angular velocity are radians per second (rad/s).  Relationship between angular and linear velocity, we again consider a pit on the rotating CD. This pit moves an arc length ∆𝑆 in a time ∆𝑡, and so it has a linear velocity ∆𝑠 𝑣= ∆𝑡 ∆𝑆 From ∆𝜃 = 𝑤𝑒 𝑠𝑒𝑒 𝑡ℎ𝑎𝑡 ∆𝑆 = 𝑟 ∆𝜃, substituting in above equation, 𝑟 𝑟∆𝜃 𝑣 𝑣= = 𝑟𝜔 or 𝜔 = ∆𝑡 𝑟 Linear velocity is proportional to the distance from the center of rotation, thus, it is largest for a point on the rim (largest r), as you might expect. 1) Microwave oven rotate at a rate of about 6 rev/min. what is this in revolutions per second? And radians per second? 2. Calculate the angular velocity of a 0.300 m radius car tire when the car travels at 15.0 m/s. Centripetal Acceleration Acceleration is a change in velocity, either in its magnitude or in its direction, or both. In uniform circular motion, the direction of the velocity changes constantly, so there is always an associated acceleration, even though the magnitude of the velocity might be constant. You experience this acceleration yourself when you turn a corner in your car. What you notice is a sideways acceleration because you and the car are changing direction. The sharper the curve and the greater your speed, the more noticeable this acceleration will become.  Figure shows an object moving in a circular path at constant speed. The direction of the instantaneous velocity is shown at two points along the path. Acceleration is in the direction of the change in velocity, which points directly toward the center of rotation (the center of the circular path). We call the acceleration of an object moving in uniform circular motion (resulting from a net external force) the centripetal acceleration(ac). Centripetal means “toward the center” or “center seeking.”  The direction of centripetal acceleration is toward the center of curvature, The magnitude of the centripetal acceleration is 𝑣2 𝑎𝑐 = 𝑟 which is the acceleration of an object in a circle of radius r at a speed v. So, centripetal acceleration is greater at high speeds and in sharp curves (smaller radius), as you have noticed when driving a car. 𝑎𝑐 in terms of angular velocity, 𝑎𝑐 = 𝑟𝜔2 3) What is the magnitude of the centripetal acceleration of a car following a curve of radius 500m at a speed of 25.0 m/s? Centripetal Force Any net force causing uniform circular motion is called a centripetal force. The direction of a centripetal force is toward the center of curvature, the same as the direction of centripetal acceleration. We know, F = ma; Newtons second law of motion. For a uniform circular motion, centripetal force is given as, 𝐹𝑐 = 𝑚𝑎𝑐 Substituting we get, 𝑣2 𝐹𝑐 = 𝑚 ; 𝐹𝑐 = 𝑚𝑟𝜔2 𝑟  Centripetal force is always perpendicular to the path and pointing to the center of curvature, because is perpendicular to the velocity and pointing to the center of curvature. 4) Calculate the centripetal force exerted on a 900 kg car that negotiates a 500 m radius curve at 25.0 m/s.  Let us now consider banked curves, where the slope of the road helps you negotiate the curve. The greater the angle θ, the faster you can take the curve. Race track for bikes as well as cars, for example, often have steeply banked curves. In an “ideally banked curve,” the angle θ is such that you can negotiate the curve at a certain speed without the aid of friction between the tires and the road. For ideal banking, the net external force equals the horizontal centripetal force in the absence of friction.  The components of the normal force N in the horizontal and vertical directions must equal the centripetal force and the weight of the car, respectively. If the angle θ is ideal for the speed and radius, then the net external force will equal the necessary centripetal force. The only two external forces acting on the car are its weight W and the normal force N of the road. Only the normal force has a horizontal component, and so this must equal the centripetal force—that is, 𝑚𝑣 2 𝑁 sin 𝜃 = 𝑟 the vertical component of the normal force is 𝑁 cos θ, and the only other vertical force is the car’s weight. These must be equal in magnitude, thus; 𝑁 cos 𝜃 = 𝑚𝑔  Combine two equations to eliminate N and get θ value as desired. Solving the second equation for N and substituting for first equation we get; sin 𝜃 𝑚𝑣 2 𝑚𝑔 = cos 𝜃 𝑟2 𝑚𝑣 𝑚𝑔 tan 𝜃 = 2 𝑟 𝑣 tan 𝜃 = 𝑟𝑔 𝑣 2 θ= tan−1 (ideally banked curve, no friction) 𝑟𝑔 4) Calculate the speed at which a 100m radius curve banked at 65o should be driven if the road is frictionless. Newton’s Universal Law of Gravitation What do aching feet, a falling apple, and the orbit of the Moon have in common? Each is caused by the gravitational force. Our feet are strained by supporting our weight—the force of Earth’s gravity on us. An apple falls from a tree because of the same force acting a few meters above Earth’s surface. And the Moon orbits Earth because gravity is able to supply the necessary centripetal force at a distance of hundreds of millions of meters. In fact, the same force causes planets to orbit the Sun, stars to orbit the center of the galaxy, and galaxies to cluster together.  Gravity is another example of underlying simplicity in nature. It is the weakest of the forces found in nature. It is a force that acts at a distance, without physical contact, and is expressed by a formula that is valid everywhere in the universe, for masses and distances that vary from the tiny to the immense. Sir Isaac Newton was the first scientist to precisely define the gravitational force, and to show that it could explain both falling bodies and astronomical motions. The gravitational force is relatively simple. It is always attractive, and it depends only on the masses involved and the distance between them.  Newton’s universal law of gravitation states that every particle in the universe attracts every other particle with a force along a line joining them. The force is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.  we assume that the body acts as if its entire mass is concentrated at one specific point called the center of mass (CM). For two bodies having masses and with a distance between their centers of mass, the equation for Newton’s universal law of gravitation is 𝑚𝑀 𝐹=𝐺 2 𝑟 where F is the magnitude of the gravitational force and G is a proportionality factor called the gravitational constant. G is a universal gravitational constant—that is, same everywhere in the universe. It has been measured experimentally to be 𝑁 𝑚2 𝐺 = 6.674 𝑥 10−11 𝑘𝑔2  In fact, our body weight is the force of attraction of the entire Earth on us with a mass of 6 x 1024 kg. The weight of an object mg is the gravitational force between it and Earth. Substituting mg for F in Newton’s universal law of gravitation gives 𝑀 𝑔=𝐺 2 𝑟 Why does Earth not remain stationary as the Moon orbits it? This is because, as expected from Newton’s third law, if Earth exerts a force on the Moon, then the Moon should exert an equal and opposite force on Earth. Tides Ocean tides are one very observable result of the Moon’s gravity acting on Earth. Because water easily flows on Earth’s surface, a high tide is created on the side of Earth nearest to the Moon, where the Moon’s gravitational pull is strongest. Why is there also a high tide on the opposite side of Earth? The answer is that Earth is pulled toward the Moon more than the water on the far side, because Earth is closer to the Moon. So, the water on the side of Earth closest to the Moon is pulled away from Earth, and Earth is pulled away from water on the far side.  As Earth rotates, the tidal bulge, keeps its orientation with the Moon. Thus, there are two tides per day (the actual tidal period is about 12 hours and 25.2 minutes), because the Moon moves in its orbit each day as well.  The Sun also affects tides, although it has about half the effect of the Moon. However, the largest tides, called spring tides, occur when Earth, the Moon, and the Sun are aligned. The smallest tides, called neap tides, occur when the Sun is at an angle 90 degree to the Earth-Moon alignment. ”Weightlessness” and Microgravity What is the effect of “weightlessness” upon an astronaut who is in orbit for months? Weightlessness doesn’t mean that an astronaut is not being acted upon by the gravitational force. There is no “zero gravity” in an astronaut’s orbit. The term just means that the astronaut is in free-fall, accelerating with the acceleration due to gravity. If an elevator cable breaks, the passengers inside will be in free fall and will experience weightlessness. You can experience short periods of weightlessness in some rides in amusement parks.  Microgravity refers to an environment in which the apparent net acceleration of a body is small compared with that produced by Earth at its surface. On Earth, blood pressure is usually higher in the feet than in the head, because the higher column of blood exerts a downward force on it, due to gravity. When standing, 70% of your blood is below the level of the heart, while in a horizontal position, just the opposite occurs. Experiments flown in space also have shown that some bacteria grow faster in microgravity than they do on Earth. Cavendish experiment The universal gravitational constant is determined experimentally. This definition was first done accurately by Henry Cavendish (1731–1810), an English scientist, in 1798, more than 100 years after Newton published his universal law of gravitation. Cavendish’s experiment was very difficult because he measured the tiny gravitational attraction between two ordinary-sized masses. Remarkably, his value for G differs by less than 1% from the best modern value.  One important consequence of knowing was that an accurate value for Earth’s mass could finally be obtained. This was done by measuring the acceleration due to gravity as accurately as possible and then calculating the mass of Earth from the relationship Newton’s universal law of gravitation gives 𝑚𝑀 mg = 𝐺 2 𝑟 m cancels and thus , 𝑀 g= 𝐺 2 𝑟 Solving for M, 𝑔𝑟 2 M= 𝐺 THANK YOU