Angular Velocity - University of Coloradojcumalat/phys1110/lectures/Lec24.pdf · Angular Velocity • Midterm on Thursday at 7:30pm! Old exams available on website. Chapters 6–9

Angular Velocity

•  Midterm on Thursday at 7:30pm! Old exams available on website. Chapters 6–9 are covered. Go to same room as last time. – You are allowed one calculator and one double-

sided sheet of paper with hand written notes – 14 Multiple choice plus two long answer questions – Test time from 7:30pm – 9:00pm. If you have a

conflict, need extra time, etc., then contact Professor Daniel Dessau.

Web page: http://www.colorado.edu/physics/phys1110/phys1110_sp12/

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Steve (m=50. kg) is skating east with velocity v=1.0 m/s . Scott (m=70. kg) is skating north with velocity v=1.0 m/s . They happen to collide at the center of the rink, and hold tight (i.e. they “stick together”, oomph!) What is the final speed of the pair, after the collision?

A) 1.4 m/s B) 1.0 m/s C) 0.5 m/s D) 0.0 m/s E) 0.72 m/s


Steve (m=50. kg) is skating east with velocity v=1.0 m/s . Scott (m=70. kg) is skating north with velocity v=1.0 m/s . They happen to collide at the center of the rink, and hold tight (i.e. they “stick together”, oomph!) What is the final speed of the pair, after the collision?

A) 1.4 m/s B) 1.0 m/s C) 0.5 m/s D) 0.0 m/s E) 0.72 m/s

€

(50kg)(1m /s)ˆ i + (70kg)(1m /s) ˆ j = (120kg) v f v f =

512

ˆ i + 712

ˆ j

€

v = (5 /12)2 + (7 /12)2 ≈ .72m /s


Four identical square tiles are glued together like a “J”, as shown. Each tile has edge length “a” and mass “m”. The origin is at the bottom right, as shown. What is the x component of the center of mass of the system of tiles?

A) -(1/2) a B) -(3/4) a C) - a D) -(3/2) a E) -3 a


Four identical square tiles are glued together like a “J”, as shown. Each tile has edge length “a” and mass “m”. The origin is at the bottom right, as shown. What is the x component of the center of mass of the system of tiles?

A) -(1/2) a B) -(3/4) a C) - a D) -(3/2) a E) -3 a

x(cm) = (m1 x1 + m2 x + m3 x3 + m4 x4 ) / (mtotal) Numbering the squares 1, 2, and 3 walking down, and #4 is the one on the left, gives x(cm) = m*(-a/2)+m*(-a/2)+m*(-a/2)+m*(-3/2a)/ (4m) = -3a/4.

A square of side 2R has a circular hole of radius R/2 removed. Relative to the center of the square the center of the hole is located at (R/2,R/2). Locate the center of mass with respect to the center of the square.

Treat the hole as negative mass: x(cm) = [4R2 (0) – π (R/2)2 (R/2)]/[4R2-π(R/2)2]

y(cm)=-.122R (same as x(cm)

Using concept of negative mass

R/2

2R

€

−πR8/[4 − π

4] = −.393R /3.21= −.122R€

CoM =mixi∑mi∑

€

CoM =msqxsq −mcirxcirmsq −mcir

Angular kinematics In chapters 2 and 3 we dealt with kinematics which involved displacement, velocity, and acceleration. Angular kinematics is the same thing but for objects which are rotating (rather than translating). For something to rotate, it must have an axis about which it rotates like the axle for a wheel.

Only sensible place for the origin is along the axis.

r θ

Use polar coordinates (r,θ) instead of Cartesian coordinates (x,y). Axis is in the z-direction.

Angular velocity Angular velocity tells us how fast (and in what direction) something is spinning.

Counterclockwise is positive

The z subscript indicates the axis is in the z-direction (and the rotation is therefore in the xy plane)

Also have angular acceleration which describes how the spinning rate changes

r θ Δθ

Using Angular Variables The angular position of a line on a disk of radius

6 cm is given by Find the average angular speed between 1 and

3 s:

€

θ =10 − 5t + 4t 2rad.

€

θ(1) =10 − 5 + 4 = 9rad.

€

θ(3) =10 −15 + 36 = 31rad.

€

ω =θ(3) −θ(1)3−1

=31− 92

=11rad /s


6 cm is given by Find the linear speed of a point on the rim at 2 s:

€

θ =10 − 5t + 4t 2rad.

€

ω =dθdt;ω(2) = −5 + 8t =11rad /s

€

v = rω = .(06m)(11rad /s)v(2) = .66m /s


6 cm is given by Find the radial and tangential acceleration of a

point on the rim at 2 s:

€

θ =10 − 5t + 4t 2rad.

€

α =dωdt

=d2θdt 2

;α(2) = 8rad /s2

€

atan = rα = .(06m)(8rad /s2)atan (2) = .48m /s2

€

arad =v 2

r=(.66m /s)2

.06marad (2) = 7.26m /s2

€

atotal = arad2 + atan

2€

dθdt

= −5 + 8t


Big Ben and a little alarm clock (synchronized to an atomic clock) both keep perfect time. Which minute hand has the largest angular velocity?

A.  Big Ben B.  little alarm clock C.  same D.  Impossible to tell


Big Ben and a little alarm clock (synchronized to an atomic clock) both keep perfect time. Which minute hand has the largest angular velocity?

A.  Big Ben B.  little alarm clock C.  same D.  Impossible to tell

Angular velocity is only a measure of how quickly the angle changes. Both minute hands complete 1 revolution every hour.

1 rph = 2π rad/hr = 2π/3600 rad/s x(cm) = (m1 x1 + m2 x + m3 x3 + m4 x4 ) / (mtotal) Numbering the squares 1, 2, and 3 walking down, and #4 is the one on the left, gives x(cm) = m*(-a/2)+m*(-a/2)+m*(-a/2)+m*(-3/2a)/ (4m) = -3a/4.

Angular velocity & acceleration vectors Note the subscript z on the angular velocity and acceleration which indicates the axis of rotation

Also, note that ω and α can be positive or negative

This is like 1D motion. Is there an equivalent 3D motion? Yes. The axis of rotation can change orientation; it is not always along the z-axis (and therefore neither is ω).

Angular velocity and acceleration are vectors: points perpendicular to the plane of rotation in the direction given by the right hand rule (direction of your thumb when fingers curl in direction of rotation).

Angular displacement Δθ measured in radians

Summary of angular kinematics r θ Δθ

Angular velocity:

Angular acceleration:

Angular velocity and acceleration are vectors:

points perpendicular to the plane of rotation in the direction given by the right hand rule (direction of your thumb when fingers curl in direction of rotation).

Angular kinematics The same equations which were derived for constant linear (or translational) acceleration apply for constant angular (or rotational) acceleration

Constant angular acceleration only!

Constant linear acceleration only!


A flywheel with a mass of 120 kg, and a radius of 0.6 m, starting at rest, has an angular acceleration of 0.1 rad/s2. How many revolutions has the wheel undergone after 10 s?

What equation should the student use to obtain the answer?

A.  B.  C.  D.  E.  None of them will work


A flywheel with a mass of 120 kg, and a radius of 0.6 m, starting at rest, has an angular acceleration of 0.1 rad/s2. How many revolutions has the wheel undergone after 10 s?

What equation should the student use to obtain the answer?

A.  B.  C.  D.  E.  None of them will work

θ0=0, ω0z=0, αz=0.1 rad/s2 and we want θ

Relationships to linear velocity If we want the linear displacement or velocity of a point on a rotating object, we need r and either θ or ω. What is the speed of the tire rim if the radius is 0.35 m and the tire rotates at 20 rad/s?

Radians measure distance around a unit circle

Therefore (only for θ in radians!)

r θ

s

Time derivative of both sides (r is constant): . Linear speed is radius times angular speed

Rim speed is

Relationships to linear acceleration What about acceleration? If speed is then

We know that centripetal (or radial) acceleration is . Using v = rω, this can be rewritten as .

Total linear acceleration is composed of both tangential and radial acceleration which are always perpendicular to each other.


A.  1 rad/s2 B.  4 rad/s2 C.  16 rad/s2 D.  1 cm/s2 E.  None of the above

5 cm CD slows from an angular velocity of 4 rad/s to a stop in 4 seconds with constant angular acceleration. What is the magnitude of the angular acceleration (α) of the CD?


A.  1 rad/s2 B.  4 rad/s2 C.  16 rad/s2 D.  1 cm/s2 E.  None of the above

5 cm CD slows from an angular velocity of 4 rad/s to a stop in 4 seconds with constant angular acceleration. What is the magnitude of the angular acceleration (α) of the CD?

Remember and for constant α we have

but want magnitude: 1 rad/s2.


A.  20 cm/s2 B.  5 cm/s2 C.  21 cm/s2 D.  0.8 cm/s2 E.  None of the above

5 cm CD has an angular acceleration of -1 rad/s2 and an angular velocity of 2 rad/s. What is the magnitude of the acceleration of the dust particle 5cm out?

Total acceleration is composed of radial and tangential acceleration.

Can now get net force by multiplying acceleration by mass

Documents

Angular Velocity - University of Coloradojcumalat/phys1110/lectures/Lec24.pdf · Angular Velocity • Midterm on Thursday at 7:30pm! Old exams available on website. Chapters 6–9