The figure below (Figure 1) illustrates an Atwood\'s machine. Let the masses of

Question

The figure below (Figure 1) illustrates an Atwood's machine. Let the masses of blocks A and B be 4.50 kg and 2.00 kg , respectively, the moment of inertia of the wheel about its axis be 0.300 kgm2 and the radius of the wheel be 0.100 m . Part A: Find the linear acceleration of block A if there is no slipping between the cord and the surface of the wheel. Part B: Find the linear acceleration 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

An atwood machine.

Mass of block A M1 = 4.50 kg

       of B M2 = 2.0 kg

The moment of inertia of the wheel about the axia = 0.300 kg.m2

radius of the wheel = 0.100 m

In this case both the block will move with the same magnitude of acceleration .

amd Acceleration of block = Angular acceleration of wheel * Radius of pulley.

Assuming value of g = 10 m /s2

Acceleration of blocks = a.

Equation of motion for A

M1*g - T1 = M1*a

or 45 - T1 = 4.5a.                 equation 1

For B

T2 - M2*g = M2*a

T2 - 20 = 2a ...                        equation 2

for wheel

Net torque = I*Angular acceleration = I(a/R)

(T1-T2)*(0.1) = 0.3*a/0.1 = 3a

0r     T1 - T2 = 30a .....                                equation 3

Solving all three equations we get

Part A

the linear acceleration of block A = 0.685 m/s2   downwards

Part B:

the linear acceleration of block  B = 0.685 m/s2   upwards

Part C:

the angular acceleration of the wheel C = a/R = 6.85 rad / s2

Part D:

the tension in left side of the cord = T1 = 41.92 N

Part E:

the tension in right side of the cord = T2 = 21.37 N

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