Physic I

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The figure is below attached.

a)

For the falling mass the free body diagram includes the weight of the mass `G =m*g (=m*a_g)` directed downwards, the string tension force `F_T` directed upwards, and the inertia of the mass itself `m*a` directed upwards (we suppose the mass is falling down).

b)

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The figure is below attached.

a)

For the falling mass the free body diagram includes the weight of the mass `G =m*g (=m*a_g)` directed downwards, the string tension force `F_T` directed upwards, and the inertia of the mass itself `m*a` directed upwards (we suppose the mass is falling down).

b)

On the vertical axis the sum of all forces need to be zero.

`F_T +m*a =m*g rArr F_T =m*(g-a)`

Of course if we impose the condition `a < <g rArr F_T ~~m*g`

c)

The string is wrapped around the spool. It means that the torque on the system is simply (the angle between the force tension in the string `F_T` and the spool radius `r_s` is 90 degree):

`T =F_T*r_s`

d)

The angular acceleration `epsilon(=alpha)` and linear acceleration `a` are related by the equation:

`a =epsilon*r_d rArr epsilon =a/r_d`

(similar to `v = omega*r_d` )

e)

The equation that relates the torque `T` , the moment of inertia `I_d` and the angular acceleration `epsilon (=alpha)` is

`T =I_d*epsilon`

(This is similar to the Newton second law `F =M*a` )

f)

For a uniform disk of radius `r_d` and mass `M_d` the moment of inertia about a perpendicular axis on disk plane (on its center) is

`I_d =(M_d*r_d^2)/2`

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