Ch 7 - Circular Motion • Circular motion: Objects moving in a circular path.
Ch 7 - Circular Motion
• Circular motion: Objects moving in a circular path.
Measuring Rotational Motion
• Rotational Motion – when an object turns about an internal axis. – Ex. Earth’s is every 24 hrs.
• Axis of rotation – the line about which the rotation occurs
• Arc length – the distance (s) measured along the circumference of the circle
• Radian- an angle whose arc length is equal to its radius, which is approximately equal to 57.3°
• When the arc length “s” is equal to the length of the radius, “r”, the angle θ swept by r is equal to one rad.
• Any angle θ is radians if defined by:
• θ = s/r
• When a point moves 360°,θ=s/r = 2πr/r = 2π radTherefore, to convert from rads to degrees
θ(rad) = π θ(deg)
180°For angular displacement,Δθ=Δs/rAngular displacement (in radians)= change in
arc length/distance from axis
Example Problem
• While riding on a carousel that is rotating clockwise, a child travels through an arc length of 11.5 m. If the child’s angular displacement is 165°, what is the radius of the carousel?
• Ans. 3.98 m
Angular Substitutes for Linear Quantities
• Linear (Straight Line)– Displacement = x– Velocity = v– Acceleration = a
• Rotational– Displacement = θ– Velocity = ω– Acceleration = α
• Angular speed – the rate at which a body rotates about an axis, expressed in radians per second
• Symbol = ω (omega) Unit = rad/s
• ω(avg) = Δθ/Δt
• ω can also be in rev/s• To convert:
1 rev = 2π rad
Example Problem
• A child at an ice cream parlor spins on a stool. The child turns counter-clockwise with an average angular speed of 4.0 rad/s. In what time interval will the child’s feet have an angular displacement of 8.0 rad
• Ans. 6.3 s
• Angular Acceleration – the time rate of change of angular speed, expressed in radians per second per second
• avg = ω2-ω1/t2-t1 =Δω/Δt
• Average angular acceleration = change in angular speed / time interval
Example Problem
• A car’s tire rotates at an initial angular speed of 21.5 rad/s. The driver accelerates, and after 3.5 s the tire’s angular speed is 28.0 rad/s. What is the tire’s average angular acceleration during the 3.5 s time interval?
• Ans. 1.9 rad/s2
Frequency vs. Period
• Frequency – # of revolutions per unit of time. Unit: revolutions/second (rev/s).
• Period – time for one revolution. Unit = second (s).
• Inversely related:
t = 1/f and f = 1/t
Tangential Velocity
• Speed that moves along a circular path.
• Right angles to the radii.
• Direction of motion is always tangent to the circle.
Rotational Speed
• The number of rotations per unit of time.
• All parts of the object rotate about their axis in the same amount of time.
• Units: RPM (revolutions per minute).
Tangential vs. Rotational
• 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.
• Tangential speed is greater on the outer edge than closer to the axis. A point on the outer edge moves a greater distance than a point at the center.
• Tangential speed = radial distance x rotational speed
v r
Centripetal Acceleration
• The acceleration of an object moving in a circle points toward the center of the circle.
• Means “center seeking” or “toward the center”. 2
c
va
r
2ca r
7.3 Forces that maintain circular motion
• Consider a ball swinging on a string. Inertia tends to make the ball stay in a straight-line path, but the string counteracts this by exerting a force on the ball that makes the ball follow a circular path.
• This force is directed along the length of the string toward the center of the circle.
The force that maintains circular motion (formerly known as centripetal force)
• Fc = (mvt2)/r
• Force that maintains circular motion = mass x (tangential speed)2 ÷ distance to axis of motion
• Fc = mrω2
• Force that maintains circular motion = mass x distance to axis x (angular speed)2
• Because this is a Force, the SI unit is the Newton (N)
Practice Problem
• A pilot is flying a small plane at 30.0 m/s in a circular path with a radius of 100.0 m. If a force of 635N is needed to maintain the pilot’s circular motion, what is the pilot’s mass?
• Answer: m = 70.6 kg
Common Misconceptions
• Inertia is often misinterpreted as a force
• Think of this example: How does a washing machine remove excess water from clothes during the spin cycle?
Newton’s Law of Universal Gravitation
• Gravitational force: a field force that always exists between two masses, regardless of the medium that separates them; the mutual force of attraction between particles of matter
• Gravitational force depends on the distance between two masses
Newton’s Law of Universal Gravitation
221
r
mGmF
Where F = ForceM1 and m2 are the masses of the two objects
R is the distance between the objectsAnd G = 6.673 x 10-11 Nm2/kg2 (constant of universal gravitation)
Practice Problem
• Find the distance between a 0.300 kg billiard ball and a 0.400 kg billiard ball if the magnitude of the gravitational force is 8.92 x 10-11 N.
• Answer: r = 3.00 x 10-1 m