tangential acceleration a [PDF]

Relating Translational and Rotational Variables Rotational position and distance moved

s = θr

(only radian units)

Rotational and → translational speed

€

 v=

ds

ds d θ = = r dt dt dt v=ωr

Relating period and rotational speed [distance = rate × time] €

T=

2 πr 2 π = v ω

ω T = 2π €

units of time (s) €

Relating Translational and Rotational Variables “acceleration” is a little tricky Rotational and translational acceleration a) from

v= ωr dv d ω = r dt dt dω at = r = αr dt

€

from change in angular speed

tangential acceleration

b) from before we know there’s also a “radial” component

€

v2 2 ar = = ω r r

radial acceleration

c) must combine two distinct rotational accelerations

€

 2  2  a tot = a r + a t 2

2

= ω 2 r + αr

2

Relating Translational and Rotational Variables Rotational position and distance moved

s = θr

(only radian units)

Rotational andtranslational speed

 dr ds d θ v= = = r dt dt dt v= ωr

€

   v =ω ×r

Rotational and translational acceleration

€

dω    r = α r tangential acceleration at = α × r dt v2 2     ar = = ω r radial acceleration a r = ω × (ω × r ) r  2  2  2 a tot = a r + a t € 2    2 a tot = a t + a r = ω 2 r + αr €

€at =

€ €

 θ (t)

   Δθ = θ 2 − θ1

(

  dθ (t) ω (t) = dt

€

€

)

€

  2  dω (t) d€ θ (t) α (t) = = dt dt 2

s = θr

   v =ω ×r

 dθ v= r = ωr dt

 a tot

2

€  2  2 = ar + at 2

= ω 2 r + αr €

€

2

   a tot = a t + a r     a r = ω × (ω × r )    at = α × r

€ €

Rotational Kinematics: ( ONLY IF α = constant)

€

ω =€ω 0 + αt θ = θ + ω 0 t + 12 αt

Example #1 A beetle rides the rim of a rotating merry-go-round. If the angular speed of the system is constant, does the beetle have a) radial acceleration and b) tangential acceleration?

If ω = constant, α = 0

 2  2  a tot = a r + a t 2

2

= ω 2r + 0

2

⇒

 a tot = ω 2 r(−rˆ ) only radial

€ If the angular speed is decreasing at a constant rate, does the beetle have a) radial acceleration and b) tangential acceleration?  v rˆ If α = neg constant, €  ω = ω 0 + αt     ar a a = a + a  2  2  2 t tot t r a tot = a r + a t € € 2 2  = ω 2 r + αr a tot € € 2 2 2 € a tot = ((ω 0 + αt) r ) +€(αr ) both radial and tangential €

Checkpoint #1 A ladybug sits at the outer edge of a merry-goround, and a gentleman bug sits halfway between her and the axis of rotation. The merry-go-round makes a complete revolution once each second. The gentleman bug’s angular speed is 1. half the ladybug’s. 2. the same as the ladybug’s. 3. twice the ladybug’s. 4. impossible to determine

Checkpoint #2 A ladybug sits at the outer edge of a merrygo-round, that is turning and speeding up. At the instant shown in the figure, the tangential component of the ladybug’s (Cartesian) acceleration is:

1.  2.  3.  4.  5.  6.  7. 

In the +x direction In the - x direction In the +y direction In the - y direction In the +z direction In the - z direction zero

Checkpoint#3 A ladybug sits at the outer edge of a merry-goround that is turning and is slowing down. The vector expressing her angular velocity is

1.  2.  3.  4.  5.  6.  7. 

In the +x direction In the - x direction In the +y direction In the - y direction In the +z direction In the - z direction zero

Problem
10‐7:


The
wheel
in
the
picture
has
a


radius
of
30cm
and
is
rota6ng
at
2.5rev/sec.
I
want
 to
shoot
a
20
cm
long
arrow
parallel
to
the
axle
 without
hi?ng
an
spokes.
(a)
What
is
the
minimum
 speed?
(b)
Does
it
maGer
where
between
the
axle
 and
rim
of
the
wheel
you
aim?



If
so
what
is
the
 best
posi6on.


The arrow must pass through the wheel in less time than it takes for the next spoke to rotate 1 rev 8 Δt= = 0.05s 2.5rev / s

(a) The minimum speed is 20cm vmn = = 400cm / s = 4m / s 0.05s

(b) No there is no dependence upon the radial position.

Problem
10‐30:
Wheel
A
of
radius
 rA=10
cm
is
coupled
by
belt
B
to
wheel
C
 of
radius
rC=25
cm.

The
angular
speed
 of
wheel
A
is
increased
from
rest
at
a
 constant
rate
of
1.6
rad/s2.
Find
the
 6me
needed
for
wheel
C
to
reach
an
 angular
speed
of
100
rev/min.
 If the belt does not slip the tangential acceleration of each wheel is the same. rAα A = rCα C

αC =

rA α A = 0.64rad / s 2 rC

The angular velocity of C is. ω C ω C rC ω C = α C t so t= = α C rAα A with ωc = 100rev/sec t=16s

Kinetic energy of Rotation TRANSLATION Review:   nd F ∝ a proportionality is inertia, m ⇒ constant of an object

Newton’s 2 Law ⇒ net energy associated with state of translational motion

€

What about rotation?

“motion” of particles with same v

1 KEtrans = mv 2 2

→ mass is translational inertia ← €

What is energy associated with state of rotational motion?

KE system = ∑ 12 m i vi2

where

= ∑ 12 m i (ωri )

2

vi = ωri

trans→ rot

= 12 [∑ ( m i ri 2 )]ω 2

€

rotational inertia (moment of inertia) about some axis of rotation

All particles have same ω

≡I

€

1

Energy of rotational motion € KE rot = 2 Iω

2

[KE

trans

= 12 mv 2 ]

Moment of Inertia I For a discrete number of particles distributed about an axis of rotation

I≡

∑m r

2

i i

units of kg·m2

all mass

Simple example:

€I = ∑ m r 4

i i

2

= m1r12 + m 2 r22 + m 2 r22 + m1r12

i=1

what about other axis?

= 2m1r12 + 2m 2 r22 - Rotational inertia (moment of inertia) only valid about some axis of rotation.

€

- For arbitrary shape, each different axis has a different moment of inertia. - I relates how the mass of a rotating body is distributed about a given axis. - r is perpendicular distance from mass to axis of rotation

Moment of inertia: comparison

I1 = ∑ m i ri 2 2

= (5 ⋅ kg)(2 ⋅ m) + (7 ⋅ kg)(2 ⋅ m) = 52 ⋅ kgm

2

I 2 = (5 ⋅ kg)(0.5 ⋅ m) 2 + (7 ⋅ kg)(4.5 ⋅ m) 2 = (1.3 + 142) ⋅ kgm 2 = 144 ⋅ kgm 2

2

Note: 5 kg mass contributes