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GRAVITATIONAL MOTION. Newton’s Law of Universal Gravitation. states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.

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## GRAVITATIONAL MOTION

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**Newton’s Law of Universal Gravitation**states that every particle in the universe attracts every other particle with a force that is directly proportional to the product of their masses and inversely proportional to the square of the distance between them.**Gravity is a consequence of mass.**The force of gravity between two objects is directly proportional to the product of the objects’ masses. The force of gravity between two objects is inversely proportional to the square of the distance between the two objects’ centers of mass.**Universal Constants**Universal Gravitational Constant: Mass of Earth: Radius of Earth:**ACCELERATION DUE TO GRAVITY**So . . . g = ? Now work it out. Now work it out. Work that problem out. Get that problem right.**Universal Gravitation #2**The mass of planet Jupiter is 1.9 x 1027 kg and that of the sun is 1.99 x 1030 kg. The mean distance of Jupiter from the sun is 7.8 x 1011m. Calculate the gravitational force which the sun exerts on Jupiter.**Torque**Is the measure of how effectively a force causes rotation.**Torques in equilibrium**When the torques associated with two masses balance each other.**Torques in equilibrium**Two people with the same mass. Which picture shows torques in equilibrium? B A**Torques in equilibrium**A – the distances are not the same in B so the torques will not balance. B A**PROBLEM 1**Alfred weighs 400 N. He sits on one end of a seesaw 1.5 m from the fulcrum. Ann weighs 200 N. How far from the fulcrum must she sit to balance the seesaw? ?**PROBLEM 1**F1 = 400 N r1 = 1.5 m F2 = 200 N r2 = ? 1 = 2 F1r1 = F2r2 400(1.5) = 200(r2) r2 = 3 m 3 m 1.5 m**Linear Motion**d – distance (in meters) v – velocity (in meters/second) a – acceleration (in meters/second2) Distance = 2r**Objects moving in a circle still have a linear velocity =**distance/time. This is often called tangential velocity, since the direction of the linear velocity is tangent to the circle. Linear/Tangential Velocity v**Angular Motion** – angular displacement (in radians) – angular velocity (in radians/second) – angular acceleration (in radians/second2) r – radius of circle (in meters)**The point or line that is the center of the circle is the**axis of rotation. If the axis of rotation is inside the object, the object is rotating (spinning). If the axis of rotation is outside the object, the object is revolving. Circular Motion Terms**Objects moving in a circle also have a rotational or angular**velocity, which is the rate angular position changes. Rotational velocity is measured in degrees/second, rotations/minute (rpm), etc. Common symbol, w (Greek letter omega) Angular Velocity Dq**If an object is rotating:**All points on the object have the same rotational (angular) velocity. All points on the object do not have the same linear (tangential) velocity. Linear velocity of a point depends on: The rotational velocity of the point. More rotational velocity means more linear velocity. The distance from the point to the axis of rotation. More distance from the axis means more linear velocity In symbols: Rotational & Linear Velocity v r w v = r w**Linear to Angular**d = r vT = r aT = r r**If an object is rotating:**All points on the object have the same Angular velocity. All points on the object do not have the same linear (tangential) velocity. Angular & Linear Velocity**Linear velocity of a point depends on:**The Angular velocity of the point. More rotational velocity means more linear velocity. The distance from the point to the axis of rotation. More distance from the axis means more linear velocity. Angular & Linear Velocity**As an object moves around a circle, its direction of motion**is constantly changing. Therefore its velocity is changing. Therefore an object moving in a circle is constantly accelerating. Acceleration**The acceleration of an object moving in a circle points**toward the center of the circle. This is called a centripetal (center pointing) acceleration. Centripetal Acceleration ac**Centripetal Acceleration**The centripetal acceleration depends on: The velocity of the object. The radius of the circle.**Circular Motion #8**A spinning ride at a carnival has an angular acceleration of 0.50 rad/s2. How far from the center is a rider who has a tangential acceleration of 3.3 m/s2?**Circular Motion #9**What is the tire’s angular acceleration if the tangential acceleration at a radius of 0.15m is 9.4 x 10-2m/s2?**Circular Motion #10**A test car moves at a constant speed of 10 m/s around a circular road of radius 50 m. Find the car’s A) centripetal acceleration and B) angular speed.**Circular Motion #11**A test car moves at a constant speed around a circular track. If the car is 48.2 m from the track’s center and has a centripetal acceleration of 8.05 m/s2, what is its tangential speed?**Circular Motion #12**A race car moves along a circular track at an angular speed of 0.512 rad/s. If the car’s centripetal acceleration is 15.4 m/s2, what is the distance between the car and the center of the track?**Circular Motion #13**A piece of clay sits 0.20 m from the center of a potter’s wheel. If the potter spins the wheel at an angular speed of 20.5 rad/s, what is the magnitude of the centripetal acceleration of the piece of clay on the wheel?**Tangential vs. Centripetal**tangential and centripetal acceleration are perpendicular to one another.**Angular Acceleration** = – angular velocity (radians/second) – angular acceleration (radians/second2) t – time (seconds)**Circular Motion #14**A ventilator fan is turning at 600 rev/min when the power is cut off, and it turns 1000 rev while coasting to a stop. Calculate the angular acceleration and the time required to stop.**Circular Motion #15**A bicycle wheel rotates with a constant angular acceleration of 3.5 rad/s2. If the initial speed of the wheel is 2 rad/s at t = 0 s. a) Through what angle does the wheel rotate in 2 s? b) what is the angular speed at t = 2 s?**Circular Motion #16**A potter’s wheel moves from rest to an angular speed of 0.20 rev/s in 30s. Find the angular acceleration in rad/s2.**Circular Motion #16**A dentist’s drill starts from rest. After 3.20 seconds of constant angular acceleration it turns at a rate of 2.51 x 104 rev/min. a) find the drill’s angular acceleration. b) Determine the angle (radians) through which the drill rotates during this period.**Circular Motion #18**A floppy disk in a computer rotates from rest up to an angular speed of 31.4 rad/s in a time of 0.892 s. A) What is the angular acceleration of the disk, assuming angular acceleration is uniform? B) How many revolutions does the disk make while coming up to speed? C) Find v if r = 4.45cm D) Find atif r = 4.45cm.

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