Lecture 16 Centripetal Acceleration ac v 2 r Circular motion at constant speed v leads to ac v 2 r It may surprise you to learn that when a mass moves around a circle of radius r at constant speed v it is still accelearting because its DIRECTION is changing This change in direction leads to a change in velocity because velocity is a vector When velocity changes there is always an acceleration given by vf vi 1 aav t By using similar triangles it is possible to show that see page 199 of text and or go to lecture v s 2 v r so that v2 v s 3 ac r t r where ac is the magnitude of the centripetal acceleration Its direction is toward the center of the circle This acceleration must be provided by a force which in the case of a mass on a string is provided by the tension in the string If this force is removed the mass goes in a straight line In the case of a car going around a corner the force is produced by the tires on the road Some notes on directions Note that centripetal acceleration ac is NOT the same as angular acceleration The angular acceleration leads to a linear acceleration a r which is TANGENTIAL to the circle Centripetal acceleration is toward the CENTER of the cirlce which is often called the radial direction The angular velocity and angular acceleration have a direction which is positive counterclockwise which you can remember by using the right hand rule Circular motion counterclockwise in the plane has a direction which is along the positive z axis The gravitational force It is worth stating again that there are only four fundamental forces The strong force and the weak force operate at length scales of order the 1 size of a nucleus 1fermi the electomagnetic force and the gravitational force dominate at larger length scales than that We have learned about forces like friction and the force on a spring Both of those forces come from electromagnetic and gravitational forces if we look closely enough The magnitude of the gravitational force between two spherical masses m1 and m2 is to a very good approximation given by F Gm1 m2 r2 4 where G 6 673 10 11N m2 kg 2 and r is the distance between the centers of the spheres The gravitational force is always attractive there is no antigravity In a famous experiment Cavandish 1798 measured G by studying the force between two masses as a function of their separation The gravitational acceleration of mass m2 due to mass m1 is a1 F m2 Gm1 r2 5 If we calculate this value for m1 mass of earth and r radius of earth then we find g 9 8m s2 Note also that Newton s third law applies here so the force on mass m1 due to mass m2 is equal and opposite to the force on m2 due to m1 Gravitational potential energy We have so far said that the change in gravitational potential energy near the surface of earth is P E mgh This is a valid equation when we consider a small mass near the surface of a large mass However there is a much more general formula given by PE Gm1 m2 r 6 The potential energy gets smaller more negative as two masses get closer together When two masses are a large distance apart the potential energy goes to zero This must reduce to mgh near the surface of a large spherical mass and it is shown in the textbook page 212 that it does Motion in a gravitational field 2 Now that we know the gravitational force and the gravitational potential we can apply Newton s second law or the work energy theory to solve a lot of problems Circular Orbits Consider motion of a planet e g earth around the sun Take the mass of the planet to be m and the mass of the sun to be M If the planet is at a distance r from the sun then the force on the planet due to the sun is the gravitational force Furthermore we can consider that the sun is stationary as its mass is so large compared to that of the planet that its motion is scarcely altered due to the force of the planet on the sun Now lets apply Newton s second law GM m ma 7 r2 What acceleration is this If the planet is moving at constant angular velocity ie 0 then the only acceleration that the earth experiences is the centripetal acceleration We can then find the velocity using the centripetal acceleration formula v2 GM m m 8 r2 r This is a very important relation because it gives a relation between the radius of the orbit and the speed at which a planet moves As we shall see later it also demonstrated Kepler s laws which were known experimentally at the time Escape velocity How fast do we need to fire a projectile mass m from the earth surface Re 6 400km Me 6 1024 kg in order for it to escape the earth s gravitational field We can solve this using the work energy theory P E KE 0 The projectile is finally at rest so KEf 0 Once the projectile has escaped it potential energy is P Ef 0 we then have P Ei KEi 0 9 Using the gravitational potential energy equation we find that 1 2 GMe m mvesc Re 2 3 10 Note that this assumes that there is no viscous drag on the projectile which in this case is a very poor approximation However we could do a similar calculation for MARS and it would be OK Anyway assuming there is no drag we get 2GMe 1 2 vesc 25 000mph 11 Re This is a lower limit on the velocity required We can also estimate the amount of energy which this requires ie 0 5mv 2 If you do the calculation it is clear that it is a large number For this reason high energy density rocket fuel is used 4
View Full Document
Unlocking...