An access turnstile is schematizable as shown. It consists of a homogeneous bar
ID: 1884396 • Letter: A
Question
An access turnstile is schematizable as shown. It consists of a homogeneous bar of negligible thickness and vertical OA of mass m1 and length L / 2, with O resting point on the ground, which is rigidly connected to another homogeneous thin bar AB of length L and mass mAB = 8m1 . A homogeneous disk of mass mD = 2 m1 of radius R, coplanar to AB, is fixed to the bar so that the axis perpendicular to the disk and passing through its center of mass is bound in B.
a) Consider the system in balance and determine the binding force and the constraining moment (independent of the previous one) exercised by the constraint in O
b) Consider therefore a rotation axis z and calculate the moment of inertia of the system with respect to this axis (the AO rod has zero thickness). The system is initially stationary and a force F is applied orthogonal to the disk and applied in B. If a braking torque is also applied M0 = -kdetermine the angular velocity of the turnstile when it is in / 2. The oscillation period is also calculated. Assume that the disk can not rotate on itself.
c) Finally calculate the kinetic energy of the system assuming that and at a certain instant it rotates with velocity 1 with respect to the ze axis, which simultaneously rotates on itself with speed 2 with respect to an axis orthogonal to the disk passing through the center of mass.
[The moment of inertia of a disk with respect to an axis passing through the C.M. and perpendicular to the face of the disk is I = 1 / 2mR2; with respect to an axis passing through the C.M. and parallel to the diameter is I = 1 / 4mR2. The moment of inertia of an auction is 1 / 12mL2 (compared to the CM). The masses, R, L, the force F and the coefficient k, in addition to the angular velocities 1 and 1 are known
a Safari 15:55 * 63% testi_appello_9_717 Home Insert Draw Layout Review View Cambria (Heading Problema 1 (punti 11) (i tre punti possono essere risolti in maniera indipendente) Un tornello di accesso è schematizzabile come in figura Esso è costituito da una sbarra omogenea di spessore trascurabile e verticale OA di massa m1 e lunghezza L/2, con O punto di appoggio al suolo, cui è collegata rigidamente un'altra sbarra sottile omogenea AB orizzontale di lunghezza L e massa mAB = 8m1, un disco omogeneo di massa mo 2 m1 di raggio R, complanare ad AB, è fissato alla sbarra in modo che l'asse perpendicolare al disco e passante per il suo centro di massa sia vincolato in B a) Si consideri il sistema in equilibrio e si determini la forza vincolare ed il momento vincolare (indipendente dalla precedente) esercitato dal vincolo in O Si consideri quindi un asse di rotazione z e si calcoli il momento di inerzia del sistema rispetto a tale asse (l'asta AO ha spessore nullo). I sistema è inizialmente fermo e viene applicata una forza F ortogonale al disco e applicata in B. Se è applicato anche un momento frenante MOF-k. determinare la velocità angolare del tornello quando si trova in -/2. Si calcoli anche il periodo di oscillazione. Si assuma che il disco non possa ruotare su se stesso Si calcoli infine l'energia cinetica del sistema assumendo che e ad un certo istante esso ruoti con velocità ! rispetto all'asse z e che contemporaneamente il disco ruoti su se stesso con velocità (02 rispetto ad un asse ortogonale al disco passante per il centro di massa b) c) [Il momento d'inerzia di un disco rispetto ad un asse passante per il C.M. e perpendicolare alla faccia del disco è 1/2mR2, rispetto ad un asse passante per il C.M. e parallelo al diametro è 1/4mR2. Il momento di inerzia di un'asta è 1/12mL2 (rispetto al CM). Siano note le masse, R, L la forza Fed il coefficiente k, oltre alle velocità angolari ()le (o 1Explanation / Answer
An access turnstile is schematizable as shown. It consists of a homogeneous bar of negligible thickness and vertical OA of mass m1 and length L / 2, with O resting point on the ground, which is rigidly connected to another homogeneous thin bar AB of length L and mass mAB = 8m1 . A homogeneous disk of mass mD = 2 m1 of radius R, coplanar to AB, is fixed to the bar so that the axis perpendicular to the disk and passing through its center of mass is bound in B.
a) Consider the system in balance and determine the binding force and the constraining moment (independent of the previous one) exercised by the constraint in O
b) Consider therefore a rotation axis z and calculate the moment of inertia of the system with respect to this axis (the AO rod has zero thickness). The system is initially stationary and a force F is applied orthogonal to the disk and applied in B. If a braking torque is also applied M0 = -kdetermine the angular velocity of the turnstile when it is in / 2. The oscillation period is also calculated. Assume that the disk can not rotate on itself.
c) Finally calculate the kinetic energy of the system assuming that and at a certain instant it rotates with velocity 1 with respect to the ze axis, which simultaneously rotates on itself with speed 2 with respect to an axis orthogonal to the disk passing through the center of mass.
[The moment of inertia of a disk with respect to an axis passing through the C.M. and perpendicular to the face of the disk is I = 1 / 2mR2; with respect to an axis passing through the C.M. and parallel to the diameter is I = 1 / 4mR2. The moment of inertia of an auction is 1 / 12mL2 (compared to the CM). The masses, R, L, the force F and the coefficient k, in addition to the angular velocities 1 and 1 are known
a. let the binding force be Fy in upward y and Fx towards roght ( +x axis)
let binding moment be M (anticlockwise)
given
mass of vertical bar = m1, length = L/2
mass of horizontal bar = 8m1, length = L
mass of disc = 2m1, radius = R
then from force balance
Fy = (m1 + 8m1 + 2m1)g = 11m1*g
Fx = 0
M = (2m1*L + 8m1*L/2 )g = 6*m1*L*g
b. moment of inertia of the system wrt z aixs = Iz
from perpendicular axis theorem moment of inertia of the disc about an axisd passing through its center of mass and paralell top z aix is
0.5*2m*R^2/2
from parallel axis theorem, about z axis it is
0.5m1*R^2 + 2m1*L^2
Iz = 8m1*L^2/3 + (0.5m1*R^2 + 2m1*L^2) = 14m1*L^2/3 + m1*R^2/2
Force applied = F
breaking torque = -k*theta
hence
from moment balance
F*L - k*theta = Iz*d(theta)/dt
integrating
t = - Iz*ln(|k*theta/F*L - 1|)/k
theta = F*L(1 + exp(-kt/Iz))/k
theta' = -F*L(exp(-kt/Iz))/Iz
theta" = FLk*exp(-kt/Iz)/Iz
hence osscilation period is T
T = 2*pi/w
w = sqrt(k/Iz)
T= 2*pi*sqrt(Iz/k)
c. KE = 0.5*Iz*w1^2 + 0.5*(2m1)*R^2*w2^2 = 0.5Iz*w1^2 + m1(R^2)*w2^2
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