Example!!#$!%$&’%(!)*!+*!’,(!-+.#)’*$(!%$&’%(!/,+0$%!+,1+0...
9
Example "#$ %$&’%( )* +* ’,( -+.#)’*$( %$&’%( /,+0$% +,1+0. %’2+2$. +2 2#$ .+3$ +*45,+% ./$$(6 7)2# 89. 2#$ .)25+:’* ). ();$%$*26 <’% + 89 2’ /,+0= 2#$ /’)*2 ’* 2#$ 89 1#$%$ 2#$ ,+.$% .#)*$. 35.2 #+>$ + ,)*$+% ./$$( ’- > ?@6AB 3C.6 D+E F. 2#$ 89 /,+0. -%’3 2#$ &$*2$% ’521+%(= (’$. )2. +*45,+% ./$$( )*&%$+.$= ($&%$+.$ ’% .2+0 &’*.2+*2G DHE <)*( 2#$ +*45,+% ./$$( ’- + 89 1#$* 2#$ ,+.$% H$+3 ). A6BI &3 -%’3 2#$ &$*2$%6 D&E J6II &3 -%’3 2#$ &$*2$%6 D(E K- 2#$ 89 /,+0. -’% JJ6B 3)*= +*( 2#$ ,+.$% H$+3 3’>$. -%’3 A6BI &3 2’ J6II &3 (5%)*4 2#). :3$= 1#+2 ). 2#$ 89L. +>$%+4$ +*45,+% +&&$,$%+:’*G ( a) Decrease (to maintain uniform linear speed!) !"#$%"&’ (b) " = v r = 1.25 0.025 = 50 rad/s (c ) " = v r = 1.25 0.06 = 20.80 rad/s ( d) " av = #$ #t = $ % $ 0 #t = 20.80 % 50 66.5 & 60 = %7.31 & 10 %3 rad/s 2 In dealing with angular velocities, displacements, be sure to keep track of the direction as we did for the linear kinematics. Constant angular acceleration problem, known : ", #, $ % $ 0 " 2 = " 0 2 + 2 # ( $ % $ 0 ) 0 2 = " 0 2 + 2 # ($2) # (20 # 2% ) )*+,-#.’ )/0.&1 2#$34.&5" 67$331+8 9"&9#$:.3 + -.;<";,+&9. =1>4 >413 -"3. =41#. 3-1&&1&0? @> >+5.3 AB ;./"#$%"&3 <"; 41, >" 9",. >" + <$## 3>"- =4.;. !"#$%!& ()*)%)&!+," 13 A ;+:C3 A - .,/ 0!12 34" &!(516 (4( 7) #, !2 27) 12!&2D M)3/,0 ’*$ ’- 2#$ 3’.2 4)N$( +2#,$2$. )* 30 /$%.’*+, >)$1 !"#$%"&’ " 0 2 = 502.64 " 0 = 22.42 rad/s 7.,+;53’ D@E 9$&$,$%+:’* ). *’2 *$4+:>$ +&&$,$%+:’* DAE 9$/$*()*4 ’* %$-$%$*&$ .0.2$3= .)4*. ’- 2#$ O5+*::$. *$$( 2’ H$ &’*.).2$*2P
Transcript of Example!!#$!%$&’%(!)*!+*!’,(!-+.#)’*$(!%$&’%(!/,+0$%!+,1+0...
Example!!"#$!%$&'%(!)*!+*!',(!-+.#)'*$(!%$&'%(!/,+0$%!+,1+0.!%'2+2$.!+2!2#$!.+3$!+*45,+%!./$$(6!7)2#!89.!2#$!.)25+:'*!).!();$%$*26!
<'%!+!89!2'!/,+0=!2#$!/')*2!'*!2#$!89!1#$%$!2#$!,+.$%!.#)*$.!35.2!
#+>$!+!,)*$+%!./$$(!'-!>!?@6AB!3C.6!!D+E!!F.!2#$!89!/,+0.!-%'3!2#$!
&$*2$%!'521+%(=!('$.!)2.!+*45,+%!./$$(!)*&%$+.$=!($&%$+.$!'%!.2+0!
&'*.2+*2G!!DHE!!<)*(!2#$!+*45,+%!./$$(!'-!+!89!1#$*!2#$!,+.$%!!
H$+3!).!A6BI!&3!-%'3!2#$!&$*2$%6!D&E!J6II!&3!-%'3!2#$!&$*2$%6!!D(E!K-!!
2#$!89!/,+0.!-'%!JJ6B!3)*=!+*(!2#$!,+.$%!H$+3!3'>$.!-%'3!A6BI!&3!!
2'!J6II!&3!(5%)*4!2#).!:3$=!1#+2!).!2#$!89L.!+>$%+4$!+*45,+%!
+&&$,$%+:'*G!
!
(a) Decrease (to maintain uniform linear speed!)
!"#$%"&'(
!
(b) " =v
r=
1.25
0.025= 50 rad/s
!
(c) " =v
r=
1.25
0.06= 20.80 rad/s
!
(d) "av =#$
#t=$ %$0
#t=
20.80 % 50
66.5 & 60= %7.31&10
%3 rad/s
2
In dealing with angular velocities, displacements, be sure to keep track of the direction as we did for the linear kinematics. !
!
Constant angular acceleration problem,
known : ", #, $ %$0
!
" 2="
0
2+ 2#($ %$
0)
!
02
="0
2+ 2 # ($2) # (20 # 2% )
)*+,-#.'!!)/0.&1(2#$34.&5"(67$331+8(
9"&9#$:.3(+(-.;<";,+&9.(=1>4(>413(-"3.(
=41#.(3-1&&1&0?((@>(>+5.3(AB(;./"#$%"&3(<";(
41,(>"(9",.(>"(+(<$##(3>"-(=4.;.(!"#$%!&'
()*)%)&!+,"'13(A(;+:C3A-''.,/'0!12'34"'&!(516'
(4('7)'#,'!2'27)'12!&2D(
M)3/,0!'*$!'-!2#$!3'.2!4)N$(!
+2#,$2$.!)*!30!/$%.'*+,!>)$1!
!"#$%"&'(
!
"0
2
= 502.64
!
"0
= 22.42 rad/s
7.,+;53'(D@E! 9$&$,$%+:'*!).!*'2!
*$4+:>$!+&&$,$%+:'*!
DAE! 9$/$*()*4!'*!
%$-$%$*&$!.0.2$3=!
.)4*.!'-!2#$!O5+*::$.!
*$$(!2'!H$!&'*.).2$*2P!!
Understanding Rotation & Circular Motion!7$!)*2%'(5&$(!$+%,)$%!2#$!*':'*!'-!2+*4$*:+,!Q!
%+()+,!+&&$,$%+:'*!-'%!!"!#$!%&"'()*%'*$+,')("-"!=!
1#+2!('$.!$+&#!&'%%$./'*(!#$%$G!
M'!%+()+,!&'3/'*$*2!'-!
,)*$+%!+&&$,$%+:'*!
!
arad =vT
2
r=(r")2
r= r" 2
M)3),+%,0=!)*!%)4)(RH'(0!%'2+:'*=!2#$!+*45,+%!
>$,'&)20!).!%$,+2$(!2'!2#$!.,!/0!-,+!&'3/'*$*2!'-!
>$,'&)20!)*!&)%&5,+%!3':'*!D1#)&#!).!./$$(!+%'5*(!
&)%&,$ES!T+4*)25($!'-!-5,,!+&&$,$%+:'*!)*!
4$*$%+,!,)*$+%!3':'*!
<)%.2=!)*!%)4)(RH'(0!%'2+:'*=!2#$!+*45,+%!
+&&$,$%+:'*!).!%$+,,0!%$,+2$(!2'!2#$!.,!/0!-,+!
&'3/'*$*2!'-!+&&$,$%+:'*!)*!&)%&5,+%!3':'*!
!
aT
= r" D(5$!2'!&#+*4)*4!+*45,+%!./$$(.E!
Differentiate using the product rule:!
Tells us how fast the !speed is changing!
Tells us how fast the !direction is changing!
Recall from an earlier lecture
8741'41'!'9,:);'<$2'27)&)'=!>'<)'=)&421'4"'2741?'
)*+,-#.(6E.&>;1<$0.8'((F,3'.2!+,,!3$()&+,!-+&),):$.!#+>$!
&$*2%)-54$.6!!"#$.$!+%$!5.5+,,0!
(%)>$*!H0!+*!$,$&2%)&!3'2'%!D',(!
'*$.!+%$!./5*!1)2#!#+*(E=!2#+2!
/52.!+*!'HU$&2!)*!%'2+:'*!+%'5*(!
+!VW$(!+W).66!!!
X+:'*+,$S!!F.+/1.;(-+;%9#.3(,"/.("$>(+#"&0(
>4.(;+:1+#(:1;.9%"&(6G"H",("<(
>$G.8I(>4.;.GJ(9+$31&0(3.-+;+%"&(
"<(>4.(-+;%9#.3(6";(G#"":(9.##38?((
K4.";.%9+##JI(9.&>;1-.>+#(<";9.3(
+;.(0;.+>.;("&(:.&3.;(60;.+>.;(
,+338(-+;%9#.3(/1+(
L(M(,(!A;((((((((((6;(M(;+:1$38(
K41&5("<(>413(+3(N310&1O9+&>#J(1&9;.+3.(>4.(<";9.("<(0;+/1>J(6,";.(3"(
&.+;(>4.(G"H",("<(>$G.8?((@,+01&.(-+;%9#.3(1&(+(P$1:(+;.(3$--";>.:(
GJ(G$"J+&9J(<";9.3?((@<(J"$(310&1O9+&>#J(1&9;.+3.(0;+/1>+%"&+#(-$##I(
3.:1,.&>+%"&(-;"9.33(=1##(4+--.&I(=4.;.(:.&3.;(6,";.(,+338(
"GQ.9>("/.;9",.3(G$"J+&9J(<";9.3(+&:(31&5(>"(>4.(G"H",?(
ExampleS!!M5//'.$!+!&$*2%)-54$!#+.!25H$.!2#+2!+%$!%'2+:*4!+2!@@=BII!%/3=!1)2#!2#$!
H'Y'3!'-!2#$!25H$.!Z6I[!&3!-%'3!2#$!+W).!
'-!2#$!%'2+:'*6!!!
D+E! <)*(!2#$!,)*$+%!./$$(!'-!2#$!H'Y'3!'-!
2#$!25H$!
DHE! 7#+2!).!2#$!&$*2%)/$2+,!+&&$,$%+:'*!+2!
2#$!H'Y'3!'-!25H$!
+W).!
&,+3/!
Z6I[!&3!
25H$!
Solution:!
!
(a) " = (11500 rev/min)#2$ rad
1 rev#
1min
60s
!
=1.2 "103 rad/s
Calculate linear speed for this rotation!
!
v = r" = 0.0907 #1.2 #103
=109 m/s
!
ac
= r" 2 = 0.0907 # 1.2 #103( )
2
=131,000 m/s2
(b) Two ways to calculate centripetal acceleration!
!
ac
=v
r
2
=109( )
2
0.0907=131,000 m/s
2 (same answer)!
Example: A catapult is used to launch a large boulder at the enemy line. Suppose the catapult arm has a length of 3 m and an angular acceleration of 5 rad/s2. What is the boulder’s linear
acceleration as a function of ! ?!
Solution: First, find the tangential component acceleration:!
!
Tangential Acceleration, aT = rd
2"
dt2
= 3# 5 =15 m/s2
"+*4$*:+,!+&&$,$%+:'*!).!*'2!2#$!'*,0!+&&$,$%+:'*!2#+2!+!H'(0!
-',,'1)*4!+!&)%&5,+%!/+2#!5*($%4'$.=!*$$(!+,.'!2#$!&'3/'*$*2!'-!
+&&$,$%+:'*!2'1+%(!&$*2$%=!
!
Centripetal (radial) Acceleration, ac =vT
2
r
Since acceleration is a vector quantity we need to consider this when determining the magnitude and direction of the acceleration of an object performing circular motion!
The centripetal (radial) component acceleration:!
!
Centripetal Acceleration, ar =vT
2
r=r"( )
2
r m/s2
!
= r" 2= r# 2
•
1$%*2)3$04-"!5))67%*7)"&).70)*"(8"!0!.)/").")9):;).7%4)%4),)$!%&"'()*%'*$+,')("-"!<)
=!4>0'5))?,!/0!-,+)*"(8"!0!.)@'0+,.0A).")40*"!A)A0'%B,-B0)"&),!/+0C)
Since both centripetal and tangential linear accelerations exist, need to treat the linear acceleration as a vector (with magnitude and direction!):!
!
Magnitude of ! a , a = atan
2+ arad
2= r " 2
+# 4
" is already given and is CONSTANT, we can use angular equivalent kinematic equations to find ! (and relate to time t which is a given):!
!
" ="0
+#t$" =#t
!
a = r " 2+" 4
t4
= r" 1+" 2t4
!
Magnitude of linear acceleration : a =15 1+ 25t 4
Sub in numbers of r=3 m, " = 5 rad/s2,!
Direction of the linear acceleration vector (relative to the centripetal one that points toward center of rotation):!
!
Direction of ! a , tan" =
aT
ac
!
= tan"1 r#
(r2$ 2
/r)
%
& '
(
) * = tan
"1 r#
(r2$ 2
/r)
%
& '
(
) * = tan
"1 #
$ 2
%
& '
(
) * = tan
"1 1
5t2
%
& '
(
) *
Energy in Rotational Motion!X'2+:'*+,!\)*$:&!]*$%40!+*(!2#$!T'3$*2!'-!K*$%:+!
'-!+!/')*2R,)^$!H'(0!
8'*.)($%!2#$!^)*$:&!$*$%40!'-!+!/')*2R,)^$!
/+%:&,$!3'>)*4!)*!+!&)%&,$!'-!%+()5.!%!
+H'52!+*!+W).6!9$V*$S!
D_3'3$*2!'-!)*$%:+`E!
"#$*!1$!&+*!1%)2$S!
Now let’s consider the same argument applied to an extended rigid body, for which all ‘points’ or mass elements of the body rotate about the fixed axis with the same angular velocity !.!
("(0!.)"&)%!0'-,)-'%!().2%)H52$(!3+..S!
"#$!H'(0!&+*!H$!+//%'W)3+2$(!+.!&'*:*5'5.S!
%'2+:'*+,!+*+,'45$!
'-!3+..6!
9)&:'*+%0!
a#0.)&+,!K*2$%/%$2+:'*!'-!2#$!T'3$*2!
'-!K*$%:+!
R$3>(+3(>4.(S1&.;%+#(,+33T(>4+>(+--.+;3(
1&(U.=>"&(@@(13(>4.(V$+&%>+%/.(
,.+3$;.("<(+(G":JT3(;.313>+&9.(>"(
$&:.;0"1&0(6#1&.+;8(+99.#.;+%"&I(>4.(
S,",.&>("<(1&.;%+T(13(>4.(V$+&%>+%/.(
,.+3$;.("<(+(G":JT3(;.313>+&9.(>"(
$&:.;0"1&0(+&(+&0$#+;(+99.#.;+%"&?(
K4.(,",.&>("<(1&.;%+(:.-.&:3(
"&((+8(>4.(,+33("<(>4.(;101:(G":JI(
G8(:13>;1G$%"&("<(>4.(,+33(+G"$>(>4.(+*13(
98(94"19.("<(+*13(+G"$>(=4194(1>(;">+>.3(
9+*&)*4!'%!.^+:*4!
!!!"(>4.(-.;3"&(=1##(3-..:($-(G.9+$3.(>4.1;(,",.&>("<(1&.;%+(4+3(:.9;.+3.:("! )W.9%/.#J(>4.(<";9.(>4.J(&..:(>"(9"$&>.;(:;+001&0(>4.(=.104>3(1&>"(>4.(9.&>;.(.*.;>3(+&:(1&>.;&+#(,",.&>(9+$31&0(>4.,(>"(4+/.(+&(+&0$#+;(+99.#.;+%"&(
!!K413(13(>4.(3+,.(+3(+&(19.(35+>.;(=4"(3-..:3($-(=4.&(34.(-$##3(1&(4.;(4+&:3(+&:(#.03(
DE008%!/)"$')!"404).").70)
/'%!A4."!0FG)
]W+3/,$S!!M5//'.$!+!4%)*(.2'*$!1)2#!+!%+()5.!'-!I6J@!3!).!H$)*4!5.$(!2'!.#+%/$*!
+!^*)-$6!!K-!2#$!,)*$+%!./$$(!'-!2#$!.2'*$!
%$,+:>$!2'!2#$!^*)-$!).!@6B!3C.=!+*(!2#$!
.2'*$L.!%'2+:'*+,!^)*$:&!$*$%40!).!@b6I!c=!
1#+2!).!)2.!3'3$*2!'-!)*$%:+G!
Solution:!
!
" =v
r=
1.5
0.61= 2.46 rad/s
First, find angular speed of grindstone!
!
K =1
2I" 2
Then, solve for Moment of Inertia from kinetic energy !
!
I =2K
" 2
!
I =2 "13.0
2.462= 4.30 kg •m2
FW).!'-!%'2+:'*!
FW).!'-!%'2+:'*!
d$*42#?AX!
E+3.(X(
E+3.(A(
]W+3/,$S!!F!(53H$,,R,)^$!($>)&$!DA!
3+..$.!!3!&'**$&2$(!H0!+!1$)4#2,$..!%'(E!
).!H$)*4!%'2+2$(!+.!&'5*2$%R&,'&^1).$6!!"#$!
%'(L.!,$*42#!).!X6!!Y4194("<(>4.(<"##"=1&0(
3>+>.,.&>(13(+G"$>(Z?[?@(13(>;$.(+&:(=4JD(
(a)!Two cases are the same!(b)! Case 1 has more!(c)! Case 2 has more!(d) Cannot be determined!
F*.1$%S!!!D&E!!!
8+.$!@S!!!K!?!3!W!XA!W!A!?!A3XA!
8+.$!AS!!!K!?!3!W!I!e!3!W!DAXEA!?!f3XA!
7"##1&0(Z"%"&(6&"(3#1-8(F!/')*2!'*!+!H)&0&,$!1#$$,!2#+2!).!)*!@$&)'
A,2!+,"!%'B,+,"!#+.!$O5+,!2+*4$*:+,!
>$,'&)20!+2!2'/!+*(!H'Y'3=!1#$%$!2#$!
%'2+:'*!+W).!).!.2+:'*+%06!!
!
vT
="r
F!/')*2!'*!+!H)&0&,$!1#$$,!2#+2!).!)*!!"!#
4+%8)'"++%!/)("-"!)7,4).,!/0!-,+)4800A)
,.).70)."8)03$,+)."5))
!
vT
= 2"r
X$+.'*)*4S!!"#$!%',,)*4!3':'*!).!+*!$;$&:>$!
&'3H)*+:'*!'-!/5%$!%'2+:'*!+*(!,)*$+%!
2%+*.,+:'*!D%'2+:'*!+W).!3'>$3$*2E6!!!
g$,'&)20!'-!%'2+:'*!+W).S!!!!!!!)
!
vT _ axis
="r
Rotation axis fixed
Rolling (rotation axis moving)
The final velocity vector at each point of a rolling Monster Truck tire
Is the vector sum of the Vresult=Vcm +Vpure (with Vpure=Vcm)
Vcm
V_pure
45
45
Vresult=0
Conservation of Energy
Kinetic Energy of Rolling Motion:!Case for YoYos or Pulleys!
!
K =1
2mv
2+1
2I" 2
!
K =1
2mv
2+1
2Iv
r
"
# $ %
& '
2
Alternative form:!
!
K =1
2mv
2+1
2mv
2 I
mr2
"
# $ %
& ' Factor out mv2:!
!
K =1
2mv
21+
I
mr2
"
# $
%
& '
It makes sense since there is energy in the rotation (as well as linear translational movement)!
