Circular Motion Tangential & Angular Acceleration

[Pages:8]Circular Motion Tangential & Angular Acceleration

? Tangential Acceleration:

vt

The arc length s is related to the angle (in radians = rad)

as follows: s = r

The tangential velocity vt is related to the angular velocity

as follows: vt = r

Tangential Velocity

The tangential acceleration at is related to the angular acceleration as follows:

at

=

dvt dt

= r d

dt

= rvt = r

= lim = d

t0 t dt

(radians/s2)

at

? Overall Acceleration:

artot = arradial + art = -aradial r^ + at^

Tangential Acceleration

Radial Axis

ar r

atot = artot =

a2 radial

+

at2

Radial Acceleration

Rick Field 2/6/2014 University of Florida

PHY 2053

Page 1

Angular Equations of Motion

? Angular Equations of Motion (constant ):

If the angular acceleration is constant then

(t) = (radians/s2)

at (t) = r

(m/s2)

(t) = 0 + t (radians/s)

vt (t) = vt0 + att (m/s)

(t)

=

0

+

0t

+

1 2

t 2

(radians)

s(t)

=

s0

+

vt 0t

+

1 2

att 2

(m)

2 (t) - 02 = 2 ( (t) -0 )

aradial (t) = r 2 (t) (m/s2)

at

v

2

t

(t

)

- vt20

=

2at (s(t)

-

s0 )

aradial (t)

=

v

2

t

(t

)

/

r

(m/s2)

Radial Axis

ar

Tangential Acceleration

r

Radial Acceleration

Rick Field 2/6/2014 University of Florida

PHY 2053

Page 2

Angular Equations of Motion

? Angular Equations of Motion (constant ):

Let N = Number of revolutions (rev) Let f = Number of revolutions per second

N

(t

)

=

(t) 2

f

(t)

=

(t) 2

(frequency)

(t) = (rad/s2)

(t) = 0 + t (rad/s)

(t)

=

0

+

0t

+

1 2

t

2

(rad)

2 (t) - 02 = 2 ( (t) -0 )

(rev/s2)

2

f (t) =

f0

+

(

2

)t

(rev/s)

N (t)

=

N0

+

f0t +

1 2

(

2

)t

2

(rev)

f

2 (t) -

f

2 0

=

2(

2

)(N (t)

-

N0 )

Rick Field 2/6/2014 University of Florida

PHY 2053

Page 3

Angular Equations: Examples

? A disk rotates about its central axis starting from rest at t = 0 and

accelerates with constant angular acceleration. At one time it is

rotating at 4 rev/s; 60 revolutions later, its angular speed is 16 rev/s.

Starting at t = 0, what is the time required to complete 64 revolutions?

2

=

f 2 (t)

2(N (t)

- -

f

2 0

N0

)

=

(16rev / s)2 - (4rev / s)2

2(60rev)

= 2rev / s2

N (t)

-

N0

=

1 2

(

2

)t 2

t = 2(N (t) - N0 ) =

(

2

)

Answer: t = 8 seconds

2(64rev) (2rev / s2 )

= 8s

? An astronaut is being tested in a centrifuge. The centrifuge has a radius R and, in starting from rest at t = 0, rotates with a constant angular acceleration = 0.25 rad/s2 . At what time t > 0 is the magnitude of the tangential acceleration equal to the magnitude of the radial acceleration (i.e. centripetal acceleration)?

aradial (t) = R 2 (t) = R 2t 2 = at = R

t= 1 =

1

= 2s

0.25rad / s2

Rick Field 2/6/2014 University of Florida

PHY 2053

Answer: t = 2 seconds

Page 4

Exam 2 Spring 2011: Problem 2

? A race car accelerates uniformly from a speed of 40 m/s to

a speed of 58 m/s in 6 seconds while traveling around a

circular track of radius 625 m. When the car reaches a

speed of 50 m/s what is the magnitude of its total acceleration (in m/s2)?

Answer: 5 % Right: 49%

at

=

v2 t2

- v1 - t1

=

(58m /

s) - (40m 6s

/

s)

=

3m /

s

ar

=

v2 R

=

(50m / s)2 625m

=

4m / s

atot = at2 + ar2 = 5m / s

Rick Field 2/6/2014 University of Florida

PHY 2053

Page 5

Gravitation: Circular Orbits (M >> m)

For circular orbits the gravitational force is perpendicular to the

velocity and hence the speed of the mass m is constant. The force

Fg is equal to the mass times the radial (i.e. centripetal) acceleration as follows:

Fg

= GmM r2

= maradial

= m v2 r

= mr 2

r

=

GM v2

(radius of the orbit, constant)

M

v

r Fg m

v = GM (speed, constant) r

Assume M >> m so that the position of M is fixed!

= GM (angular velocity, constant) r3

For circular orbits r, v, and are also constant.

T = 2r = 2r r = 2 r3

v

GM

GM

(period of rotation)

? Kepler's Third Law:

T 2 = 4 2r 3

GM

The period squared is proportional to the radius cubed.

Rick Field 2/6/2014 University of Florida

PHY 2053

In general both masses rotate about the center-of-mass and the formulas are more complicated!

M

CM

? rM VM

vm Fg m rm

Page 6

Circular Orbits: Example

? Two satellites are in circular orbit around the Earth. The first satellite

has mass M1 and is travelling in a circular orbit of radius R1. The second satellite has mass M2 = M1 is travelling in a circular orbit of radius R2 = 4R1. If the first satellite completes one revolution of the Earth in time T, how long does it take the second satellite to make one

revolution of the Earth?

Answer: 8T

T12

=

4 2 R13

GM 1

T22

=

4 2 R23

GM 2

=

4 2 (4R1)3

GM1

= 64 4 2R13

GM 1

= 64T12

T2 = 8T1 = 8T

Rick Field 2/6/2014 University of Florida

PHY 2053

Page 7

Circular Orbits: Example

? Two diametrically opposed masses m revolve around a circle of radius R. A third mass M = 2m is located at the m center of the circle. What is the period T of rotation for this system of three masses? Answer: T = 4 R3

3 Gm

M = 2m

m

R

( ) Fgrav

=

GmM R2

+ Gm2 (2R)2

= Gm R2

M

+

1 4

m

= maradial

= m v2 R

m F M = 2m

m

R

v=

G(M

+

1 4

m)

R

T = 2R = 2R

R

= 2R

R

= 2R 4R = 4 R3

v

G(M

+

1 4

m)

G(2m

+

1 4

m)

9Gm 3 Gm

Rick Field 2/6/2014 University of Florida

PHY 2053

Page 8

 ................
................

In order to avoid copyright disputes, this page is only a partial summary.

Google Online Preview   Download