Question 2: For a 1-kg ball being pushed by a rod in the vertical path with equation r = 0.50, where is in radian and r is in meter, at the instant shown, if it has = 2 rad/s and Ö = 4 rad/s², determine the force the rod exerts on the ball. Neglect friction. (a) 10.28 N (b) 12.66 N (c) 15.73 N (d) 9.81 N O r 0= R|N π 2

Elements Of Electromagnetics
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ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
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Question 2: For a 1-kg ball being pushed by a rod in the vertical
path with equation r = 0.50, where is in radian and r is in meter, at
the instant shown, if it has = 2 rad/s and = 4 rad/s², determine
the force the rod exerts on the ball. Neglect friction.
(a) 10.28 N
(b)
12.66 N
(c) 15.73 N
(d)
9.81 N
O
r
Ꮎ ;
RIN
π
Transcribed Image Text:Question 2: For a 1-kg ball being pushed by a rod in the vertical path with equation r = 0.50, where is in radian and r is in meter, at the instant shown, if it has = 2 rad/s and = 4 rad/s², determine the force the rod exerts on the ball. Neglect friction. (a) 10.28 N (b) 12.66 N (c) 15.73 N (d) 9.81 N O r Ꮎ ; RIN π
Ex For a lkg ball being pushed by a rod in the
vertical path with equation r = 0.50 where is
in radian and r is in meter.. at the instant shown
if it has = 2 rad/s and 8 = 4 rad/s², determine.
it ragial and transverse accelerations
Y=0.50
r=0.50
r=0.50
10
2
r N
at o
W
0 = 11/1/201
ar = r-r0
= 2 - 0.7853/2)²
-1.1412 m/s2
00=r0+2+0 = 0.7853 (4) +2(1)(2) = 7.1412 m/s²
r=0.5 (2) = 0.7853
r=0.5 (2)
= 1
r=0.5(4)
=
Y=tan!
2
r
dr/do
=
tan 0.7853= 57.5151
0.5
tarigent Line
+
← FO=F-Ncos 57.5151, - mao
=
F= 12.409 N
Transcribed Image Text:Ex For a lkg ball being pushed by a rod in the vertical path with equation r = 0.50 where is in radian and r is in meter.. at the instant shown if it has = 2 rad/s and 8 = 4 rad/s², determine. it ragial and transverse accelerations Y=0.50 r=0.50 r=0.50 10 2 r N at o W 0 = 11/1/201 ar = r-r0 = 2 - 0.7853/2)² -1.1412 m/s2 00=r0+2+0 = 0.7853 (4) +2(1)(2) = 7.1412 m/s² r=0.5 (2) = 0.7853 r=0.5 (2) = 1 r=0.5(4) = Y=tan! 2 r dr/do = tan 0.7853= 57.5151 0.5 tarigent Line + ← FO=F-Ncos 57.5151, - mao = F= 12.409 N
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