Discussion 3: TAM212, Fall, 2003
Name: solutions
The box in the figure travels along the industrial conveyor. If a box (mass = 1kg) pictured to the right starts from rest at A, and the belt increases its speed such that at = (0.2t) m/s2, where t is

Discussion 7: TAM212, Spring, 2003
Name: solutions
The gear of radius R rolls on the horizontal rack. Pin G in the center of the gear engates a slot in the arm AB, & which rotates at the constant angular velocity = counterclockwise as shown. Dete

Discussion 8: TAM212, Spring, 2003
Name:
j
FA G
mg
k i
T
The vertical bar is fixed. The other bar AB (of uniform mass m and length L) is initially held at the angle shown. It is attached to a collar that can slide vertically without friction on

4. If we let "W" refer the cylindrical wheel, "C" refer to its center, and "I" where it touches the surface, and we let "R" refer to the rod, setting i to be the direction of the motion of the wheel, j, directed from "C" to "I", and (finally) let "F"

Discussion 9: TAM212, Spring, 2003
Name: A 1 slug homogeneous cylindrical disk is given the counterclockwise velocity of 4 rad/s with the spring unstretched. The spring constant, k, equals 3 lb/ft. If the disk does not slip while rolling, how far wi

Extra problem from Exam 2 TAM212, Spring, 2002 1.
Name: Section:
A cylinder of radius R = 10cm rolls on the horizontal surface shown. A pin in the center of the cylinder engages a slot in the bar AB which is pinned at point A. Bar AB rotates with a

TAM 212 Spring, 2004, Exam 1A
1.
Name: solutions
& = Fnormal = ma normal Fnormal sin = ms 2 /
FWeight
F F
horizontal
Fnormal sin = mv 2 / R
(1) (2)
FNormal
(1) (2) tan = v 2 / Rg = tan -1 (v 2 / Rg )
Vertical
= maVertical Fnormal

TAM 212 Spring, 2003 Exam 2 Solutions 1. Fext dt = L(t 2 ) - L(t1 ) = mv c (t 2 ) - mv c (t1 )
t1 t2 = t t2
t1 = 0 t2 = t
Fext dt = 0 mi v i (t ) = mi v i (0) mv = (m + 2m )v f v f = v / 3
i i t2 = t i i t1 = 0 i i
2. Fext dt = mi v i

y
TAM 212, Spring, 2004, solutions
5d
1. xOC =
c c I zz = I zz c zz rod
( ) + (I ) ; ) = (W / g )L / 12 + (W / g )(L / 2) = (1 / 3)(W / g )L = (1 / 3)(W / g )(10d ) (I ) = (I ) + m (r (r ) = (2 / 5)(W / g )R + (W / g )R = (7 / 5)(W / g )25d = (35)

TAM 212, Spring, 2004, Exam 2B solutions 1. Just before to just after the collision, e is defined by e = (v R 2 - v P 2 ) (v P1 - v R1 ) . Using work energy leading rest, - (1 / 2 )m R v R 2 = - m R gH v R 2 = - 2 / 5 gH (where the positive and nega

TAM 212, Spring, 2004, Exam 2a solutions v 1. Using v R / F = v R / B + v O '/ F + B rO ' R , with v R / F = -vi ; v v R / B = v R / B 2 2 (i + j) ; v O '/ F = 0 ; B = B k , and rO 'P = d (i + j) ,
B
R
j
k
v
i
(
-vi = v R / B = vR / B
(

3.1
Given: Pictures with geometry shown Find: Which objects are in simple plane motion Solution: a, c, d, and e are all in plane motion. A good indication is that the rotational axis i.e. there is a point of zero velocity for each plane of rotation

TAM212 A1) vC = 10i C k r I C = 10i
Hour Exam II - Spring 2005
Solution- Version A
C k 1j = 10i -C i = 10i C = -10 rad/s
vA = C k r I A = -10k 2j = 20i m/s
vB vB i vB i vB i j components: 0 B i components: vB vB vB
= = = = = = = = =
vA +

Name
SS#
Section
TAM212 - Hour Exam 1 - Spring 2005
Put your name on each sheet of work Show all work on these exam sheets Please write clearly and legibly
Formula Sheet:
Cylindrical coordinates: vp = rer + r e + zk 2 ap = r - r er + r

Name
SS#
Section
TAM212 - Hour Exam 1 - Spring 2005
Put your name on each sheet of work Show all work on these exam sheets Please write clearly and legibly
Formula Sheet:
Cylindrical coordinates: vp = rer + r e + zk 2 ap = r - r er + r

Name
SS#
Section
TAM212 - Hour Exam 1 - Spring 2005
Put your name on each sheet of work Show all work on these exam sheets Please write clearly and legibly
Formula Sheet:
Cylindrical coordinates: vp = rer + r e + zk 2 ap = r - r er + r