1 Circular Motion

the motion or spin on an internal axis

the motion or spin on an external axis

Number of rotations per unit of time Rpm or Rps All objects that rotated on same axis have the same rotational speed. Also called Frequency (cycles/s or Hertz) Period (seconds) is the inverse of Frequency

7 Gymnast on a High Bar A gymnast on a high bar swings through two rotations or cycles in a time of 1.90s. Find the average rotational speed (in rps) or frequency (in Hz) of the gymnast.

Given: t = 1.90 s & 2 rotation (cycle) Find the average rotational speed (in rps) Rps= rotations second second = 2 rotation 1.90 seconds = 1.05 rps = 1.05 cycles/second

9 A Helicopter Blade A Helicopter Blade Find the rotational speed or frequency at #1 if it takes s for one rotation (cycle)?

Given: t = s & 1 rotation (cycle) Find the average rotational speed (in rps) Rps= rotations second second = 1 rotation seconds = 6.49 rps = 6.49 cycles/second

11 A Helicopter Blade A Helicopter Blade Find the rotational speed or frequency at #2 if takes s for one rotation (cycle)?

Given: t = s & 1 rotation (cycle) Find the average rotational speed (in rps) Rps= rotations second second = 1 rotation seconds = 6.49 rps = 6.49 cycles/second

Do Frequency/ Rotational Speed Problems

The speed in m/s of something moving along a circular path. It always tangent to the circle.

The distance moved per unit of time.The distance moved per unit of time. Linear speed is greater on the outer edge of a rotating object than it is closer to the axis.Linear speed is greater on the outer edge of a rotating object than it is closer to the axis. Linear Speed Linear Speed

2πr Distance traveled in one period is the circumference 2πr T Time for one “cycle” is the “period” (T) Tangential Speed = Circumference / Period Tangential Speed = 2πr T

But remember that period is the inverse of frequency So instead of dividing by period you multiply by frequency Tangential Speed = Circumference x Frequency Tangential Speed 2πr x cycle Tangential Speed = 2πr x cycle cycles cycle s

18 A Helicopter Blade A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 1 on the blade, find the tangential speed

Given: r = 3.00 m & Angular speed = 6.50 rps Angular speed = 6.50 rps Tangential Speed = 2πr x cycle cycle s cycle s = 2π 3.00m x 6.50 cycle cycle s cycle s = 122 m/s

20 A Helicopter Blade A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 2 on the blade, find the tangential speed

Given: r = 6.70 m & Angular speed = 6.50 rps Angular speed = 6.50 rps Tangential Speed = 2π 6.70m x 6.50 cycle cycle s cycle s = 273 m/s Notice that the tangential speed at 3 meter is 122 m/s while at 6.70 meters is 273 m/s

Do Tangential Speed Linear Velocity Problems

23 Centripetal Acceleration (centripetal acceleration)

24 A Helicopter Blade A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 1 on the blade, find the tangential acceleration

Given: r = 3.00 m & Tangential Speed = 122 m/s Tangential Speed = 122 m/sTangential Acceleration = (122 m/s) 2 / 3.00m = 4,960 m/s 2 = 4.96 x 10 3 m/s 2

26 A Helicopter Blade A Helicopter Blade A helicopter blade has an angular speed of 6.50 rps. For points 2 on the blade, find the tangential acceleration

Given: r = 6.70 m & Tangential Speed = 273 m/s Tangential Acceleration = (273 m/s) 2 / 6.70m = 11,200 m/s 2 = 1.12 x 10 4 m/s 2

Do Centripetal AccelerationProblems

Centripetal Force F c = ma c F c = mv T 2 r

Centrifugal force: Center-fleeing, away form center

Vertical drum rotates, you’re pressed against wall Friction force against wall matches gravity Seem to stick to wall, feel very heavy The forces real and perceived Real Forces: Friction; up Centripetal; inwards Gravity (weight); down Perceived Forces: Centrifugal; outwards Gravity (weight); down Perceived weight; down and out

Weight the force due to gravity on an object Weight = Mass  Acceleration of Gravity W = m g Weightlessness - a conditions wherein gravitational pull appears to be lacking Examples: Astronauts Falling in an Elevator Skydiving Underwater

From 2001: A Space Odyssey rotates like bicycle tire Just like spinning drum in amusement park, create gravity in space via rotation Where is the “floor”? Where would you still feel weightless? Note the windows on the face of the wheel

Do Centripetal Force Problems

What makes something rotate? TORQUE AXLE How do I apply a force to make the rod rotate about the axle? Not just anywhere!

Torque = force times lever arm Torque = F  L

Torque example F L What is the torque on a bolt applied with a wrench that has a lever arm of 30 cm with a force of 30 N? Torque = F x L = 30 N x 0.30 m = 30 N x 0.30 m = 9 N m = 9 N m For the same force, you get more torque with a bigger wrench  the job is easier!

Net Force = 0, Net Torque ≠ 0 10 N > The net force = 0, since the forces are applied in opposite directions so it will not accelerate. opposite directions so it will not accelerate..

Net Force = 0, Net Torque ≠ 0 10 N > However, together these forces will make the rod > However, together these forces will make the rod rotate in the clockwise direction. rotate in the clockwise direction.

Net torque = 0, net force ≠ 0 The rod will accelerate upward under these two forces, but will not rotate.

Balancing torques 10 N 20 N 1 m 0.5 m Left torque = 10 N x 1 m = 10 n m Right torque = 20 N x 0.5 m = 10 N m

Balancing torques 10 N 20 N 1 m 0.5 m Left torque = 10 N x 1 m = 10 n m Right torque = 20 N x 0.5 m = 10 N m How much force is exerted up by the Fulcrum?

Torque = force times lever arm Torque = F  L

Equilibrium To ensure that an object does not accelerate or rotate two conditions must be met: To ensure that an object does not accelerate or rotate two conditions must be met:  net force = 0  net torque = 0

Example 1 Given M = 120 kg. Neglect the mass of the beam. Find the Torque exerted by the mass Torque = F L = 120 kg (9.8 m/s 2 ) (7 m) = 120 kg (9.8 m/s 2 ) (7 m) = 8232 N m

Example Given: W box =300 N W box =300 N Find: F TR = F CC = ? N ACDACD Torque C = Torque CC F C L = F CC L 300 N (6 m) = F CC (8 m) 300 N (6 m) = F CC (8 m) 225 N = F CC

Example Given: W box =300 N W box =300 N Find: F TL = F CC = ? N ACDACD Torque C = Torque CC F C L = F CC L F C (8 m) = 300N (2 m) F C (8 m) = 300N (2 m) F C = 75 N

Example ACDACD Does this make sense? F TL = 75 N F TR = 225 N Does the F UP = F DOWN ? F UP = 75 N N = 300 N =F DOWN(Box) F UP = 75 N N = 300 N =F DOWN(Box) Given: W box =300 N W box =300 N

Another Example Given: W=50 N, L=0.35 m, x=0.03 m Find the tension in the muscle x L W Torque C = Torque CC F C L = F CC L 50N (0.350 m) = F CC (0.030m) 50N (0.350 m) = F CC (0.030m) 50N (0.350 m) / (0.030m) = F CC 583 N = F CC

Stability CM &Torque

Condition for stability If the CG is above the edge, the object will not fall CG

when does it fall over? CG STABLE NOT STABLE If the vertical line extending down from the CG is inside the edge the object will return to its upright position  the torque due to gravity brings it back.

Stable and Unstable stable unstable torque due to gravity pulls object back torque due to gravity pulls object down

Stable structures Structures are wider at their base to lower their center of gravity

If the center of gravity is supported, the blocks do not fall over Playing with your blocks CG

Rotational Inertia The rotational “laziness” of an object

Recall : inertia A measure of the “laziness” of an object because of Quantified by the mass (kg) of object

Rotational Inertia (I) A measure of an object’s “laziness” to changes in rotational motion Depends on mass AND distance of mass from axis of rotation

Balancing Pole increases Rotational Inertia

Angular Momentum

Momentum resulting from an object moving in linear motion is called linear momentum. Momentum resulting from the rotation (or spin) of an object is called angular momentum.

Conservation of Angular Momentum Angular momentum is important because it obeys a conservation law, as does linear momentum. The total angular momentum of a closed system stays the same.

Calculating angular momentum Angular momentum is calculated in a similar way to linear momentum, except the mass and velocity are replaced by the moment of inertia and angular velocity. Angularvelocity(rad/sec) Angularmomentum (kg m/sec 2 ) L = I w Moment of inertia (kg m 2 )

Gyroscopes Angular Momentum A gyroscope is a device that contains a spinning object with a lot of angular momentum. Gyroscopes can do amazing tricks because they conserve angular momentum. For example, a spinning gyroscope can easily balance on a pencil point.

A gyroscope on the space shuttle is mounted at the center of mass, allowing a computer to measure rotation of the spacecraft in three dimensions. An on-board computer is able to accurately measure the rotation of the shuttle and maintain its orientation in space. Gyroscopes Angular Momentum

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