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PS 11 GeneralPhysics I for the Life Sciences

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PS 11 GeneralPhysics I for the Life Sciences

Rotational Motion

Dr. Benjamin chan

Associate Professor

Physics Department

february 2014

- 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

- 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

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

- 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?

- 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

- 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

- Average a
- Instantaneous a
- Make Dt as small as possible

- Tangential acceleration
- Radial acceleration

- Frequency = number of complete revolutions per second = f
- w = 2pf

- Period = time required to complete one revolution = T = 1/f

- 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

- 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

- 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!

- 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?

- 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)

- The number of bits passing the head per second is

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

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

- aR
- Variable magnitude, variable direction

- 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

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

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

- 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.

- 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

- 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).

- 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

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)

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

- 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

- a) Straight leg

units

- 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!

- For a height of 172 cm

units

units

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

Center of mass of wrench follows constant velocity trajectory

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

- Units: Nm (Newton-meter)
- Reserve J for work and energy

- Torque is a vector quantity
- Direction determined by the right hand rule

- Translational Equilibrium
- All forces cancel out: SF = 0

- Rotational Equilibrium
- Torques must balance out: SG = 0

- When is an object in equilibrium?

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

- 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

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

Solution

It increases by

- 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

- 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.

- 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)

- 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!!

- 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.

- W = FDl = F rDq
- W = tDq
- Power
P = W/Dt

P = tDq/Dt = tw

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

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

- How can the ice skater spin so fast?

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

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

- What happens when you tilt the rotating disk?

HINT:

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

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.

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

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