Let the masses of blocks A and B be 4.50kg and 2.00kg , respectively, the moment

Question

Let the masses of blocks A and B be 4.50kg and 2.00kg , respectively, the moment of inertia of the wheel about its axis be 0.400kg?m2, and the radius of the wheel be 0.130m . Find the linear accelerations of block A if there is no slipping between the cord and the surface of the wheel. Find the linear accelerations of block B if there is no slipping between the cord and the surface of the wheel. Find the angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel. Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel. Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel. A string is wrapped several times around the rim of a small hoop with radius 8.00 cm and mass 0.180kg . The free end of the string is held in place and the hoop is released from rest . After the hoop has descended 95.0cm , calculate the angular speed of the rotating hoop and the speed of its center.

Explanation / Answer

net Torque = I*alfa

Ta*R - Tb*R = I*alfa

(Ta-Tb)*R = I*a/R

(Ta -Tb)*R^2 = I*a

Ta - Tb = 23.67a..........(1)

along vertical

Ma*g - Ta = Ma*a.........(2)

Tb - Mb*g = Mb*a...........(3)

2+3 gives


Tb - Ta + Ma*g - Mb*g = (Ma+Mb)*a.........(4)

substituting 1 in 4

- 23.67a + (4.5-2)*9.8 = (2+4.5)*a


a = 0.812 m/s^2 down wards



B) a = 0.812 m/s^2 upwards

C) alfa = a/R = 6.246 rad/s^2

D) from 3
Tb - Mb*g = Mb*a

Tb = 21.224 N

E) Ma*g - Ta = Ma*a



Ta = 40.446 N


2) I = M*R^2 = 0.18*0.08^2 = 0.001152 kg m^2

M*g*h = 0.5*I*W^2 = 0.5*M*R^2*W^2

g*h = 0.5*R^2*W^2

W = 53.93 rad/s

V = R*W = 4.3144 m/s

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