Atwood\'s machine problem? Let the masses of blocks A and B be 4.50 kg and 2.00

Question

Atwood's machine problem? 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 0.400 kg*m^2 and the radius of the wheel be 0.150 m.
Block B is lower than block A in the picture.

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

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

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

(D) Find the tension in right side of the cord. >>> I can do this if someone can show me the formula they used for part C)
Thank you!

Explanation / Answer

call T1 the tension in the rope connected to mass 1 (mass A); T2 is the tension in the other rope

newton's second laws for the masses are

T1 - m 1 g = - m1a (the negative sign occurs because m1, the heavier mass, will descend and down
is the negative direction)

T2 - m2 g = + m2 a

for the wheel, it experiences a net torque of (T1-T2)R because the tensions are different on its two sides

this new torque causes an angular acceleration given by

(T1-T2)R = I alpha where I is themoment of inertia of the disk and alpha the angular acceleration

but, for a wheel, I = 1/2 MR^2 and angular accel = a/R where R is the radius of the wheel and M its mass

therefore (T1-T2)R = 1/2 MR^2 *(a/R) => T1-T2 = 1/2 Ma

subtract the first two equations and get

T1 - T2 - m1g + m2g = - m1 a - m2 a

group and rearrange: a = (m1-m2)g/[1/2 M + m1 +m2]

you are given m1 and m2, find M knowing I = 0.4 = 1/2 MR^2 and R = 0.14m

substitute into the equation and solve for a

angular accel = a/R

find the tensions by using this value of a in the first two equations for the masses (T1-m1g = - m1a and T2- m2g = m2a) to solve for T1 and T2

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