Respuesta :

2g/3r is the  magnitude α of the angular acceleration of the cylinder as the block descends.

mg - T = ma

T = tension in rope

a = acceleration of block

Torque balance around cylinder

Torque = Iα= Tr

Icylinder = 0.5mr^2

α= a/r

0.5mr^2a/r = Tr

T = 0.5ma

mg - T = ma

mg - 0.5ma = ma

mg = 1.5ma

a = 2g/3

α= a/r = 2g/3r

The rate at which the angular velocity in a circular motion changes over time is known as the angular acceleration. Rotational acceleration is another name for it. Given that it has both a magnitude and a direction, it is a vector quantity. Angular acceleration is represented by the letter alpha.

Variable velocity is what causes the acceleration. The phrase "angular acceleration" is used to describe changes in rotational speed in rotating objects. According to the connection, the angular displacement of a rotating object depends on time t. Every time a body rotates, it experiences angular acceleration. A quantitative vector can be used to express the change in angular velocity per unit of time.

To know more about   angular acceleration visit : brainly.com/question/14769426

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