Chapter 8. Centripetal Force and The Law of Gravity

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1 Chapter 8 Centripetal Force and The Law of Gravity

2 Centripetal Acceleration An object traveling in a circle, even though it moves with a constant speed, will have an acceleration The centripetal acceleration is due to the change in the direction of the velocity The tangential component of the acceleration is due to changing speed

3 Centripetal Acceleration, cont. Centripetal refers to center-seeking The direction of the velocity changes The acceleration is directed toward the center of the circle of motion

4 Centripetal Acceleration and Angular Velocity The angular velocity and the linear velocity are related (v v = ωr) The centripetal acceleration can also be related to the angular velocity a 2 C = ω r Newton s 2nd Law says that the centripetal acceleration is accompanied by a force F = ma C F stands for any force that keeps an object following a circular path, such as tension in a string, gravity, friction

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6 Applications of Forces Causing Centripetal Acceleration Many specific situations will use forces that cause centripetal acceleration Level curves Banked curves Horizontal circles Vertical circles

7 Level Curves Friction is the force that produces the centripetal acceleration Can find the frictional force, µ, v 2 mv F fr = = µ mg r v = µrg

8 Banked Curves A component of the normal force adds to the frictional force to allow higher speeds mg = n cosθ Ffr = nsinθ = mg tanθ F fr 2 mv = = mg tanθ r tan θ = v 2 rg

9 Horizontal Circle The horizontal component of the tension causes the centripetal acceleration a C = g tanθ

10 Vertical Circle Look at the forces at the top of the circle The minimum speed at the top of the circle can be found v top = gr

11 Newton s Law of Universal Gravitation Every particle in the Universe attracts every other particle with a force that is directly proportional to the product of the masses and inversely proportional to the square of the distance between them. F = G m m 1 r 2 2

12 Law of Gravitation, cont. G is the constant of universal gravitational G = x N m² /kg² This is an example of an inverse square law

13 Applications of Universal Gravitation Mass of the earth Use an example of an object close to the surface of the earth r ~ R E M E = gr G 2 E

14 Applications of Universal Gravitation Acceleration due to gravity g will vary with altitude g = G M r E 2

15 Gravitation Constant Determined experimentally Henry Cavendish 1798 The light beam and mirror serve to amplify the motion

16 Chapter 5 Statics

17 Equilibrium An object either at rest or moving with a constant velocity is said to be in equilibrium The net force acting on the object is zero F = 0 In terms of its components: F = 0 F = 0 x y

18 Solving Equilibrium Problems Make a sketch of the situation described in the problem Draw a free body diagram for the isolated object under consideration and label all the forces acting on it Resolve the forces into x- and y-components, using a convenient coordinate system Apply equations, keeping track of signs Solve the resulting equations

19 Equilibrium Example

20 Torque Torque, τ, is the tendency of a force to rotate an object about some axis τ = Fd τ is the torque F is the force d is the lever arm (or moment arm) Units (SI) Newton x meter = Nm

21 Direction of Torque Torque is a vector quantity The direction is perpendicular to the plane determined by the lever arm and the force For two dimensional problems, into or out of the plane of the paper will be sufficient If the turning tendency of the force is counterclockwise, the torque will be positive If the turning tendency is clockwise, the torque will be negative

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23 Lever Arm The lever arm, d, is the perpendicular distance from the axis of rotation to a line drawn from the axis of rotation to a line drawn along the the direction of the force d = L sin Φ

24 An Alternative Look at Torque The force could also be resolved into its x- and y- components The x-component, F cos Φ,, produces 0 torque The y-component, F sin Φ,, produces a non-zero torque

25 Torque, final From the components of the force or from the lever arm, τ = FL sinφ F is the force L is the distance along the object Φ is the angle between the force and the object