Problem 10.63: Atwood\'s machine. The figure below (Figure 1) illustrates an Atw

Question

Problem 10.63: Atwood's machine.

The figure below (Figure 1) illustrates an Atwood's machine.

Part A

Let the masses of blocks A andB be 4.50kg and 2.00kg , respectively, the moment of inertia of the wheel about its axis be 0.300kg?m2, and the radius of the wheel be 0.110m . Find the linear accelerations of block A if there is no slipping between the cord and the surface of the wheel.

Part B

Find the linear accelerations of block  B if there is no slipping between the cord and the surface of the wheel.

Part C

Find the angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel.

Part D

Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel.

Part E

Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.

Problem 10.63: Atwood's machine. The figure below (Figure 1) illustrates an Atwood's machine. Part A 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 0.300kg?m2, and the radius of the wheel be 0.110m . Find the linear accelerations of block A if there is no slipping between the cord and the surface of the wheel. Part B Find the linear accelerations of block B if there is no slipping between the cord and the surface of the wheel. Part C Find the angular acceleration of the wheel C if there is no slipping between the cord and the surface of the wheel. Part D Find the tension in left side of the cord if there is no slipping between the cord and the surface of the wheel. Part E Find the tension in right side of the cord if there is no slipping between the cord and the surface of the wheel.

Explanation / Answer

Part(A)

torque at pulley = r x Ta - r x Tb = I*alpha     and alpha = a/r

Ta - Tb = Ia/r^2 = 24.79a   .....(i)

at Block B:

Tb - Ma*g = (Ma)a

Tb - 2g = 2a .... (ii)

On block A :

4.50g - Ta =4.50a    ...(iii)

add (i), (ii) and (iii) ,

4.50g - 2g = (24.79 + 2+4.5)a

a = 0.784 m/s2

B) a = 0.784 m/s2

C) alpha = a/r = 0.784/0.110 = 7.12 rad/s2

D) Tb = 2(g +a)   = 2(0.784+9.81) = 21.19 N

E) Ta = 4.50 ( g-a) = 40.62 N

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