- By
**lam** - Follow User

- 81 Views
- Uploaded on

Download Presentation
## PowerPoint Slideshow about ' PS 11 GeneralPhysics I for the Life Sciences' - lam

**An Image/Link below is provided (as is) to download presentation**

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -

Presentation Transcript

### PS 11 GeneralPhysics I for the Life Sciences

Rotational Motion

Dr. Benjamin chan

Associate Professor

Physics Department

february 2014

Questions and Problems for Contemplation

- Chapter 8
- Questions: 1, 4, 8, 10, 15, 18, 22, 24
- Problems: 1, 2, 4, 8, 10, 15, 22, 27, 30, 38, 45, 52, 55, 61, 64
- General Problems: 72, 80, 81

Describing Rotational Motion

- Angular displacement q
- Let O be the axis of rotation
- How far the object has rotated
- Only 2 directions possible: clockwise(-) and counter-clockwise(+)
- Measured in radians
- 1 radian (rad) is the angle subtended by an arc whose length is equal to the radius of motion

Distance traveled

- Arc length traversed
- For one complete revolution
- q can be expressed in revolutions

Example: Bike Wheel

- A bike wheel rotates 4.5 revolutions. How many radians has it rotated?
- If the wheel has a diameter of 45 cm, what is the distance traveled by a point on the rim of the wheel?

Example: Bird of Prey

- A bird’s eye can distinguish objects that subtend an angle no smaller than 3 10-4rad. How many degrees is this?
- How small an object can the bird just distinguish when flying at a height of 100m?
- For small angles (<15), arc length and chord length are nearly the same

Angular Velocity w

- Average w
- Instantaneous w
- Dt must be very small
- Velocity v of a point on a rotating wheel
- Changes direction as vector turns
- Increases in proportion to distance from the axis of rotation

Angular Acceleration a

- Average a
- Instantaneous a
- Make Dt as small as possible
- Tangential acceleration
- Radial acceleration

Review of Linear and Angular Quantities

- Frequency = number of complete revolutions per second = f
- w = 2pf
- Period = time required to complete one revolution = T = 1/f

Equations of Motion

- Zero angular acceleration
- a = 0, w = constant
- Uniform circular motion
- q = wt + qo
- Linear velocity is not constant
- Magnitude is constant: v = wr
- Direction is changing
- Acceleration is not constant
- atan= 0 but aR= rw2 = constant (centripetal)
- Direction is changing

Example: Earth’s Rotation

- How fast is the earth’s equator turning?
- w = 2p/T = (2prad)/84,600s = 7.27 x 10-5rad/s
- v = rw = (6,380 km)(7.27 x 10-5rad/s) = 464 m/s
- How will your speed change as you go to the North or South pole?
- v = (r cosf)w = (464 cosf) m/s
- f = 14.5°, v = 449 m/s
- f = 30°, v = 402 m/s
- f = 60°, v = 232 m/s
- f = 90°, v = 0 m/s

The Coriolis Effect

- As you go from the equator towards the N pole, you are moving faster than the ground you are moving into: veer to your right (earth rotates west to east)
- As you go from the N pole towards the equator, you are moving slower than the ground you are moving into: veer to your right
- Clockwise flow!
- To or from the S pole: veer to the left!
- Counterclockwise flow!

Example: Hard Drive

- The platter of the hard drive of a computer rotates at 7200 rpm. What is the angular velocity of the platter?
- If the reading head of the drive is 3.00 cm from the axis of rotation, how fast is the disk moving right under the head?

Example (continued)

- If a single bit requires 0.50 mm of length along the direction of motion, how many bits per second can the writing head write when it is 3.00 cm from the axis?
- The number of bits passing the head per second is

or 45 megabits/s (Mbps)

Constant Angular Acceleration

- a = constant
- w = wo + at
- q = qo + wot + ½ at2
- Eliminate t between w and q
- w2 = wo2 + 2aq

Total Acceleration

- atotal = atan + aR
- atan
- Constant magnitude, changing direction
- aR
- Variable magnitude, variable direction

Example: Centrifuge

- A centrifuge motor is accelerated from rest to 20,000 rpm in 30s. Determine its angular acceleration and how many revolutions it makes while it is accelerating.
- Solution
- Assuming constant angular acceleration

Example continued

- Where the final angular velocity w is
- The angular displacement in 30s is then
- We divide by 2p to convert to revolutions

Rolling Motion

- Translational + rotational motion
- No Slipping
- Static friction between object and rolling surface

Example: Bicycle

- A bicycle slows down uniformly from a velocity of 8.40 m/s to rest over a distance of 115 m. The overall diameter of the tire is 68.0 cm. Determine the initial angular velocity of the wheels.

Example (continued)

- Determine the number of revolutions each wheel undergoes before stopping.
- The rim of the wheel turns 115m before stopping. Thus,
- Determine the angular acceleration of the wheel

Example continued

- Determine the time it took the bicycle to stop
- Note: when the bike tire completes one revolution, the bike advances a distance equal to the outer circumference of the tire (no slipping or sliding).

Announcements

- FINAL EXAM
- Wednesday, March 19
- 7.30 - 10.30
- F-113
- Long Test 4
- Thursday, March 13
- 6.00 – 7.30
- Room TBA c/o Paulo

Center of Mass

You can reduce an object to a point and describe its translational motion by considering the motion of this point (called its center of mass)

Determining Center of Mass

Consider masses m1, m2, m3, … with coordinates (x1, y1), (x2, y2), (x3, y3), …

CM for a Leg

- Determine the center of mass of a leg when a) stretched out and b) bent at 90°. Assume the person is 1.70 m tall.
- Solution
- a) Straight leg
- Essentially 1-D
- Measure distance from hip joint
- CM is 52.1-20.4 = 31.7 units from base of foot
- For a height of 172 cm, xcm = 54.5 cm above the bottom of the foot

units

CM of Leg

- b) Bent leg
- For a height of 172 cm

xcm = (172 cm)(0.149) = 25.6 cm

ycm = (172 cm)(0.23) = 39.6 cm

- Center of mass of bent leg is 39.6 cm above the floor and 25.6 cm from thehip joint!

units

units

CM Trajectory

Center of mass of swimmer in flight follows projectile motion (parabolic) trajectory

Center of mass of wrench follows constant velocity trajectory

Torque

- What causes an object to rotate?
- Torque = force x lever arm

More Torque

- Units: Nm (Newton-meter)
- Reserve J for work and energy
- Torque is a vector quantity
- Direction determined by the right hand rule

Newton’s First law

- Translational Equilibrium
- All forces cancel out: SF = 0
- Rotational Equilibrium
- Torques must balance out: SG = 0
- When is an object in equilibrium?

Newton’s Second Law

- F = ma
- G = Ia
- I = moment of inertia
- a = angular acceleration
- Only two possible directions
- Counter-clockwise rotation
- Clockwise rotation

Moment of Inertia of Particles

- For a single moving object with mass m
- t = rF = rma = rmra =mr2a
- I = mr2
- For several objects rigidly attached to each other
- St = (Smiri2)a
- I = Smiri2

Changing Moment of Inertia

Determine the change in the moment of inertia of a particle as the radius of its orbit doubles

Solution

It increases by

Changing Your I

- Vertical axis of rotation
- Arms on the side
- R = 25 cm, M = 9.6 kg
- Raise your arms in a crucifixion pose
- R = 57.5 cm, M = 9.6 kg
- 433% increase

Example: Ball Rolling Down an Inclined Plane

- Determine the speed of a solid sphere of mass M and radius R when it reaches the bottom of an inclined plane if it starts from rest at a height H and rolls without slipping. Assume no slipping occurs. Compare the result to an object of the same mass sliding down a frictionless inclined plane.

Solution

- Initial mechanical energy
- PE = MgH
- KEtrans = 0
- KErot = 0
- Final mechanical energy
- PE = 0
- KEtrans = ½ Mv2
- KErot= ½ Iw2
- Conservation of Energy

MgH = ½ (Mv2 + Iw2)

Solution (continued)

- I = (2/5)MR2 for a solid sphere rotating about an axis through its center of mass
- w = v/R
- Thus

MgH = ½ Mv2+ ½(2/5)MR2(v/R)2

(1/2 + 1/5) v2 = gH

v = [(10/7)gH]1/2

- v does not depend on the mass and radius of the sphere!!

Frictionless Incline

- Ball slides down the incline and does not roll
- Thus,

½ Mv2 = MgH

v = (2gH)1/2

- The speed is greater!
- None of the original PE is converted into rotational energy.

Angular Momentum L

- L = Iw
- Newton’s second law becomes
- Thus,

Conservation of Angular Momentum

- If the net torque acting on a rotating object is zero, then its angular momentum remains constant.

The Ice Skater

- How can the ice skater spin so fast?

The Diver

- How can the diver make somersaults?
- Does she have to rotate initially?
- What trajectory does she follow?

The Hanging Wheel

- Why is the wheel standing up?
- Why does it turn around about the point of support?

Drunk Driver Test/Tightrope Artist

- Follow the line walk
- Increase your moment of inertia to minimize rotations

Quiz 8

1. A 4 kg mass sits at the origin, and a 10 kg mass sits at x = + 21 m. Where is the center of mass on the x-axis?

(a) + 7 m (b) + 10.5 m (c) + 14 m (d) + 15 m

2. An object moving in a circular path experiences

(a) free fall.

(b) constant acceleration.

(c) linear acceleration.

(d) centripetal acceleration.

3. A boy and a girl are riding on a merry-go-round which is turning at a constant rate. The boy is near the outer edge, and the girl is closer to the center. Who has the greater angular velocity?

a) The boy

b) The girl

c) Both have the same non-zero angular velocity.

d) Both have zero angular velocity.

Quiz 8

4. A wheel starts at rest, and has an angular acceleration of 4 rad/s2. Through what angle does it turn in 3 s?

a) 36 rad b) 18 rad c) 12 rad d) 9 rad

5. A wheel of diameter 26 cm turns at 1500 rpm. How far will a point on the outer rim move in 2 s?

a) 314 cm b) 4084 cm c) 8995.5 cm d) 17990.8 cm

6. What is the centripetal acceleration of a point on the perimeter of a bicycle wheel of diameter 70 cm when the bike is moving 8 m/s?

a) 91 m/s2 b) 183 m/s2 c) 206 m/s2 d) 266 m/s2

7. A bicycle is moving 4 m/s. What is the angular speed of a wheel if its radius is 30 cm?

a) 0.36 rad/s b) 1.2 rad/s c) 4.8 rad/s d) 13.3 rad/s

Quiz 8

8. An ice skater is in a spin with his arms outstretched. If he pulls in his arms, what happens to his kinetic energy?

a) It increases. b) It decreases.

c) It remains constant but non-zero. d) It remains zero.

9. What is the quantity used to measure an object\'s resistance to changes in rotation?

a) mass b) moment of inertia

c) linear momentum d) angular momentum

10. A wheel of moment of inertia of 5.00 kg-m2 starts from rest and accelerates under a constant torque of 3.00 N-m for 8.00 s. What is the wheel\'s rotational kinetic energy at the end of 8.00 s?

a) 57.6 J b) 64.0 J c) 78.8 J d) 122 J

Download Presentation

Connecting to Server..