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Chapter 7: Rotational Motion and the Law of Gravity. A unit of angular measure: radian. y. Angular Speed & Acceleration. length of the arc from the x-axis s:. P. s = r q where s,r in m, and q in rad(ian) . r. q. A complete circle: s = 2 p r. 360 o = 2 p rad. x.
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Chapter 7: Rotational Motion and the Law of Gravity • A unit of angular measure: radian y Angular Speed & Acceleration length of the arc from the x-axis s: P s = rq where s,r in m, and q in rad(ian) r q A complete circle: s = 2pr 360o = 2p rad x 57.3o= 1 rad
Momentum and Impulse • Angular displacement and velocity y Angular displacement: Dq = qf - qi in a time interval Dt = tf – ti P at tf Average angular velocity: r rad/s P at ti qf Instantaneous angular velocity: qi x rad/s w < (>) 0 (counter) clockwise rotation
Momentum and Impulse • Angular acceleration y Average angular acceleration: rad/s2 P at tf Instantaneous angular acceleration: r rad/s2 P at ti qf qi x
Momentum and Impulse • An example of a rigid body • The distance of any two points in a • rigid object does not change when • the body is even in motion. • When a rigid object rotates about • a fixed axis, every portion of the object • has the same angular speed and the • same angular acceleration.
Rotational Motion under Constant Angular Acceleration • One-to-one correspondence between linear and angular quantities • Similarity between wav and vav • Similar derivation used for linear quantities can be used for • angular quantities.
Rotational Motion under Constant Angular Acceleration • An example • Example 7.2 : A rotating wheel A wheel rotates with a constant angular acceleration of 3.50 rad/s2. If the angular speed of the wheel is 2.00 rad/s at ti=0, (a) through what angle does the wheel rotate between t=0 and t=2.00 s? (b) What is the angular speed of the wheel at t=2.00 s?
Relations between Angular and Linear Quantities • w and v • Consider an object rotating about • the z-axis and a point P on it. tangent to circle tangential speed The tangential speed of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular speed.
Relations between Angular and Linear Quantities • a and a • Consider an object rotating about • the z-axis and a point P on it. tangent to circle tangential acceleration The tangential acceleration of a point on a rotating object equals the distance of that point from the axis of rotation multiplied by the angular acceleration.
Centripetal Acceleration • Acceleration at a constant speed • Consider a car moving in a circular • path with constant linear speed v. Even though the magnitude of vi and vf are the same, Dv can be non-zero if their directions are different. This leads to non-zero acceleration called centripetal acceleration.
Centripetal Acceleration • Centripetal acceleration • Consider a car moving in a circular • path with constant linear speed v. • Triangle OAB and the triangle • in Fig. (b) are similar. Total acceleration
Centripetal Acceleration • Vector nature of angular quantities • Angular quantities are vector and their directions are defined as: w points out of the page w points into the page
Centripetal Acceleration • Forces causing centripetal acceleration • An object can have a centripetal acceleration only if some external • force acts on it. • An example is a ball whirling in a circle at the end of a string. In this • case the tension in the string is the force that creates the centripetal • force. Net centripetal force Fc is the sum of the radial components of all forces acting on a given object. T=Fc • A net force causing a centripetal • acceleration acts toward the center • of the circular path. If it vanishes, • the object would immediately leave • its circular path and move along a • straight line tangent to the circle.
Centripetal Acceleration • Examples • Example 7.7 : Buckle up for safety Find the minimum coefficient of static friction ms between tires and roadway to keep the car from sliding. v=13.4 m/s r=50.0 m
Centripetal Acceleration • Examples • Example 7.8 : Daytona International Speedway (a) Find the necessary centripetal acceleration on the banked curve so that the car will not slip due to the inclination (neglect friction). y-component (vertical) : q=31.0o r=316 m x-component (horizontal) : (b) Find the speed of the car.
Centripetal Acceleration • Examples • Example 7.9 : Riding the tracks (a) Find the speed at the top. R=10.0 m (b) Find the speed at the bottom.
Centripetal Acceleration • Examples • Example 7.9 : Riding the tracks (cont’d) • Find the normal force on a passenger • at the bottom if R=10.0 m n does not depend on R!
Newtonian Gravitation • Law of universal gravitation • Using accumulated data on the motions of the Moon and planets, • and his first law, Newton deduced the existence of the gravitational • force that is responsible for the movement of the Moon and planets • and this force acts between any two objects. If two particles with mass m1 and m2 are separated by a distance r, then a gravitational force acts along a line joining them with magnitude Newton’s 2nd law constant of universal gravity
Newtonian Gravitation • Law of universal gravitation (cont’d) • The gravitational force exerted by a • uniform sphere on a particle outside the • sphere is the same as the force exerted • if the entire mass of the sphere were • concentrated at its center. This is a result from Gauss’s law and stems from the fact that the gravitational force is inversely proportional to square of the distance between two particles. • The expression F=mg is valid only • near the surface of Earth and can be • derived from Newton’s law of universal • gravitation.
Newtonian Gravitation • Gravitational potential energy revisited • Gravitational potential energy near Earth (approximation) • General form of gravitational potential energy due to Earth mass of Earth radius of Earth This is a special case where the zero level for potential energy is at an infinite distance from the center of Earth. The gravitational potential energy associated with an object is nothing more than the negative of the work done by the force of gravity in moving the object.
Newtonian Gravitation • Gravitational potential energy revisited (cont’d) • Derivation of gravitational potential energy near Earth
Newtonian Gravitation • Escape speed • If an object is projected upward from Earth’s surface with a large • enough speed, it can soar off into space and never return. This • speed is called Earth’s escape speed vesc. The initial mechanical energy of the object-Earth system is: If we neglect air resistance and assume that the initial speed is large enough to allow the object to reach infinity with a speed of zero, this value of vi is the escape speed vesc. 4.3 km/s for Mercury 11.2 for Earth 2.3 for Moon 60.0 for Jupiter
Newtonian Gravitation • Examples m1,2,3 =0.300 kg • Example 7.10 : Billiards • Find the net gravitational force • on the cue ball.
Newtonian Gravitation • Examples m1,2,3 =0.300 kg • Example 7.10 : Billiards (cont’d) (b) Find the components of the force of m2 on m3.
Newtonian Gravitation • Examples m1,2,3 =0.300 kg • Example 7.12 : A near-Earth asteroid • An asteroid with mass m=1.00x109 kg comes from infinity, and falls • toward Earth. • Find the change in potential energy when it reaches a point 4.00x108 • m from Earth. Find the work done by gravity. ri=0. (b) Find the speed of the asteroid when it reaches rf=4.00x108 m. (c) Find the work needed to reduce the speed by half.