A thin uniform disk of mass M =5.00kg and radius R =0.40m has its center of mass

Question

A thin uniform disk of mass M=5.00kg and radius R=0.40m has its center of mass is attached a distance L/4 from the bottom end of a uniform rod of mass m=2.50kg and length L=1.60m The top end of the rod is attached to a frictionless pivot. The system is displaced 90o with respect to the vertical and released from rest.

A thin uniform disk of mass M=5.00kg and radius R-0 40m has its center of mass is attached a distance -from the bottom end of a uniform rod of mass m and length L-1.60m The top end of the rod is attached to a frictionless pivot. The system is displaced 900 with respect to the vertical and released from rest. sig (a) Calculate the location of the center mass of the system relative to the fixed end of the rod (b) Calculate the moment of inertia of the system. kgm2 (c) Calculate the angular speed of the system when =30 x rad/s (d) Calculate the centripetal acceleration of the center of mass of the system when =300. (e) Calculate the tangential acceleration of the center of mass of the system when theta-30 (f) Calculate the magnitude of the force at the pivot when theta- 30° (g) Calculate the direction of the force at the pivot of the when theta-30° X m/s2 X m/s2 X (measured counterclockwise with respect to the positive x-axis)

Explanation / Answer

(A) r_cm = (m1 r1 + m2 r2)/(m1 + m2)

= [ (5 x 3L/4) + (2.50 x L/2)]/(5 + 2.50)

= 0.667L

= 1.07 m

(B) I1 = (2.50 x 1.60^2 / 12) = 0.533 kg m^2

I2 = (5 x 0.4^2 / 2) + (5 x (3 x 1.60/4)^2) )

I = I1 + I2 = 8.13 kg m^2

(C) Applying energy conservation,

(5 + 2.50) g 1.07 (1 - cos30) = 8.13 w^2 / 2

w = 1.61 rad/s

(d) a_c = w^2 r = 2.78 m/s^2

(e) 7.50 x 9.81 x 1.07 x sin30 = 8.13 alpha

alpha = 4.84 rad/s^2

a_t = alpha r = 5.18 m/s^2

(f) F - (5+2.50)(9.81)j = (7.50)[ (-2.78sin30i + 2.78 cos30j) + (4.84)(-cos30i - sin30j)]

{ Applying Fnet = m a on center of mass}

F = - 41.9i + 73.5j

magnitude = sqrt(41.9^2 + 73.5^2) = 84.6 N

(g) theta = 180 - tan^-1(73.5/41.9)

= 120 deg

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