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Transcript
Lecture 4.4 Applications of Newton’s Second Law Circular Motion We have already talked about various applications of the Newton’s second law. However, the only examples which we considered were examples of translational motion, while studying kinematics we also learned how to describe rotational motion. Now we shall look at rotational dynamics. Think what you know about acceleration of the point-like object moving around a circle. When we discussed kinematics of rotation, we saw that rotation is always accompanied with acceleration, known as centripetal acceleration. We proved that it is always directed towards the center of the circle. Indeed in order to be able to make the turn a particle should change the direction of its velocity towards the center of the arc path, so the acceleration should be directed in a same way. The magnitude of this acceleration is acp v2 . r (4.4.1) 1. Centripetal force As we learned from studying the laws of Newtonian mechanics, acceleration can not come from nowhere. If a body has acceleration then, according to the Newton's second law, the nonzero net force should exist which produces this acceleration. Since we are talking about centripetal acceleration, we shall call this force a centripetal force. It is directed to the center of the circle and has the magnitude of v2 F m . r (4.4.2) I deliberately have not used any special letter for centripetal force, because there is no any special centripetal force in nature. Indeed equation 4.4.2 is just another way of writing Newton's second law for the case of the circular motion. This means that significance of the force in the left-hand side of this equation is just the net force acting on the body. We have already discussed that the net force is not a special force but just the result of the vector addition of all the forces acting on the body. In a same way the centripetal force is the result of the vector addition of all the forces acting on the body. You do not have to show centripetal force as an extra force in the free body diagram. However, if the body moves around a circle then it has centripetal acceleration directed to the center of this circle. You have to take this into account when performing vector addition of the forces. This is why it is convenient to choose one of the coordinate axes in the direction of the radius of the circle and another in the direction perpendicular to the radius (tangential direction). Let us consider several simple examples. The passenger is sitting at the back seat of the car which makes a turn. This passenger can move along the seat if he/she is not holding or using the belt. This is because the passenger experiences the effect of inertia. Indeed according to Newton's first law this passenger should continue to move along the straight line if no forces are acting on him/her. In order to be able to make this turn he/she must experience a force which will provide the centripetal acceleration. In the case of the car it is either the force of the belt or the force of friction between the passenger and the seat. As you can see, it is not some special centripetal force, but just one of the regular forces acting on this person. Another good example would be motion of the Moon around the Earth. Considering Moon's orbit as a circle, we have to find the centripetal force which is responsible for its circular motion. The only force acting on the Moon is gravitational force of the Earth. It is directed to the center of the Moon's orbit. Even though it is centripetal force, but it has common gravitational nature. If it is not this force acting on the Moon, it will just fly away from the Earth along the straight line (according to the Newton's first law). Sometimes people ask a question why the Moon is not falling down to the Earth if there is gravitational attraction between them. This is because the Moon has finite orbital velocity. In a sense it is falling, but it also moves in tangential direction, so finally it just makes one more turn around the Earth. If something suddenly stops the Moon then it will fall down to the Earth. Example 4.4.1. The coefficient of static friction between the road and the tires of the car is . What speed will put the car on the verge of sliding as it rounds a leveled curve of radius R? When the car makes this turn it should have centripetal acceleration directed along the ground to the center of the curve. The only force which may act on the car in this direction is the force of friction. It is static friction, since the car does not slide from the curve. So the force of static friction is the centripetal force for this case and we have m v2 Ffs , R The maximum value of this static friction force is Ffs max N mg . Since for the car is on horizontal road then N mg . This gives 2 vmax m mg , R 2 vmax Rg , vmax Rg , which is the maximum possible speed for this car before it slides from the curve.