Circular Motion

The Radian Objects moving in circular (or nearly circular) paths are often measured in radians rather than degrees. In the diagram, the angle θ, in radians, is defined as follows So, if s = r the angle is 1 rad and if s is equal to the full circumference of the circle, the angle is 2π rad. (In other words, 360° = 2π rad.)

Radians and Degrees

Angular Displacement (θ) If we consider a small body to be moving round the circle from A to B we say that it has experienced an angular displacement of θ radians. The relation between the (linear) distance moved, d, of the body and the angular displacement θ, is given by Also, if the angle is small, d is very nearly equal to the magnitude of the linear displacement of the body. d = rθ

Angular Velocity (ω) Suppose that the body moved from P 1 to P 2 in a time t. The linear speed, v, of the body is given by v = d/t. If we divide d=rθ by t, we have The angular velocity ω is defined as Units of ω are rad/s Therefore, v = rω

Merry-Go Round What happens to your speed as you go to the middle of a merry-go round? a. The speed remains constant. b. The speed increases c. The speed decreases

Merry-Go Round What happens to your angular speed as you go to the middle of a merry- go round? a. The angular speed remains constant. b. The angular speed increases c. The angular speed decreases

Angular Acceleration(α) Previously we assumed that the body moved from P 1 to P 2 with constant speed. If the linear speed of the body changes then, obviously, the angular speed (velocity) also changes. The angular acceleration, α, is the rate of change of angular velocity. So, if the angular velocity changes uniformly from ω 1 to ω 2 in time t, then we can write: Now, linear acceleration, a, is given by Substituting v=rω  We find,

Circular Motion Definitions Time Period, T The time period of a circular motion is the time taken for one revolution. Rotational Frequency, f The rotational frequency of a circular motion is the number of revolutions per unit time. Time period is the inverse of frequency Also,and

Rotational Motion What is the relationship between Liner and Rotational motion quantities?

Acceleration Equations Revisited

What is the Acceleration? a. No acceleration b. Acceleration outward c. Acceleration toward the center of the circle d. Acceleration points tangential to the circle A plane attached to a string flies in a circle at constant speed Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

What is the Acceleration? speed velocitynot Even though the speed is constant, velocity is not constant since the direction is changing:  acceleration! If released the plane would travel at a constant speed tangentially. Therefore, the change in velocity would be a vector from A to P If θ is very small, this acceleration (Δv/t) points to the center of the circle. This is called centripetal acceleration Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Centripetal Acceleration speedvelocity not Even though the speed is constant, velocity is not constant since the direction is changing: must be some acceleration! Consider average acceleration in time Δt a As we shrink Δ t, Δv / Δt dv / dt = a

Centripetal Acceleration Magnitude: r Direction:- r (toward center of circle) Since and v = ωr points toward the center of the circle a = ω2r Since and v = ωr a = ω 2 r

Centripetal Acceleration

Centripetal Acceleration (a c ) ‘Centripetal’ means center seeking Vector direction always points toward the center of the circle Magnitude: v = speed r = radius of the circle

Centripetal Acceleration (a c )

What is the Net Force? a. No net force b. Net Force points outward c. Net Force points toward the center of the circle d. Net Force points tangential to the circle A plane attached to a string flies in a circle at constant speed Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

What is the Net Force? a. No net force b. Net Force points outward c. Net Force points toward the center of the circle d. Net Force points tangential to the circle A plane attached to a string flies in a circle at constant speed Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

What is the Net Force? Centripetal Force, F c, is this net force. Recall, Newton's 2nd Law F net =ma Therefore, F c = ma c A plane attached to a string flies in a circle at constant speed Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Centripetal Force (F c ) acts toward the center of the circle depends on mass, speed, and size of the circle Net force, provided by another force or interactions of forces

Date Physics

Centripetal Acceleration Example When you are driving a car, and you turn the steering wheel sharply to the right in order to turn the car to the right, you "feel" as if a force is pushing your body to the left against the door. In order for your body to follow the car in the tight circular path, something has to push your body toward the center of the circle-- in this case it is the driver's-side door-- and your tendency otherwise is to travel in a straight line tangent to the circular path

Centripetal Force Source What provides the centripetal force (F c ) in the following scenarios? A car going around a corner at a constant speed. Friction force between the tires and the pavement Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Centripetal Force Source What provides the centripetal force (F c ) in the following scenarios? A ball on a string being twirled in a circle. Tension force of the string on the ball nd/lawsCentripetalForce_files/image004.jpg

Centripetal Force Source What provides the centripetal force (F c ) in the following scenarios? The planets orbiting around the sun. The sun’s force of gravity on the planets.

Centripetal Force Source What provides the centripetal force (F c ) in the following scenarios? A motorcycle stuntman going around a loop. Both the normal force of the track and gravity. Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Centripetal Force Example A ball attached to a string is twirled in a circle of radius 0.5 m with a speed of 2 m/s. If the ball has a mass of 0.25 kg, what is the tension of the string. Knowns: m = 0.25 kg v = 2 m/s r = 0.5 m T = 2 N

The normal reaction, F N, has no component acting towards the center of the circular path. Therefore the required centripetal acceleration is provided by the force of friction, F f, between the wheel and the road. Moving in a Straight line on a Horizontal Surface If the force of friction is not strong enough, the vehicle will skid.

Turning on a Banked Surface The normal reaction, F N, now has a component acting towards the center of the circular path. If the angle, , is just right, the correct centripetal acceleration can be provided by the horizontal component of the normal reaction. This means that, even if there is very little force of friction the vehicle can still go round the curve with no tendency to skid. Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Angle of Banking The magnitude of the horizontal component of the normal force is This force causes the centripetal acceleration, so, the magnitude of N X is also given by So, The vertical forces acting on the vehicle are in equilibrium. Therefore, summing the vertical forces Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Angle of Banking Solving for F N and substituting intogives: Simplifying, This equation allows us to calculate the angle θ needed for a vehicle to go round the curve at a given speed, v, without any tendency to skid. Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

NASCAR Physics What is the minimum speed a NASCAR car must go not to slide down the banked curves of the Bristol Motor Speedway if the banked angle is 36° and the radius is 73.5m?

Loopy The minimum speed to complete a loop requires: Speed large enough to reach the top of the loop. At the top of the loop F net = F g + F N For the minimum speed F N = 0 Therefore, recall So, at the top of the loop Cutnell & Johnson, Wiley Publishing, Physics 5 th Ed.

Kepler’s Law of Periods Proof Assumptions: Must conform to equations for circular motion Newton’s Universal Law of Gravity Planet rotates in a circular (elliptical) path Newton’s 3 rd Law  symmetry Recall,so Therefore, rearranging Law of Periods

Testing the Inverse Square Law of Gravitation The acceleration due to gravity at the surface of the earth is 9.8m/s 2. If the inverse square relationship for gravity (F g ~1/r 2 ) is correct, then, at a distance ~60 times further away from the center of the earth, the acceleration due to gravity should be The centripetal acceleration of the moon is given by where the radius of the moon’s orbit is r = 3.84 × 10 8 m and the time period of the moon’s orbital motion is T = 27.3 days.