Figure 2 shows a constant force, f of magnitude 20n pulled a light cord wrapped around a pulley to lift a bucket of mass 1.53kg. the radius and the moment of inertia of the pulley are 0.330m and 0.385kg m² respectively. if the frictional torque is 1.10nm calculate the
(a) angular acceleration of the pulley. (b) linear acceleration of the bucket

(a)

The angular acceleration of the pulley can be found by using the equivalent of Newton's second law for rotational motions:

(1)

where

is the net torque on the pulley

is the moment of inertia of the pulley

is the angular acceleration

First, we need to find the net torque. The torque exerted by the force F (forward) is:

While the frictional torque (backward) is . So, the net torque is

Now, re-arranging eq.(1), we find the angular acceleration:

(b)

Assuming that the string holding the bucket is inextensible, the bucket should have the same linear acceleration of a point on the edge of the pulley, which is given by

where

is the angular acceleration

r = 0.330 m is the radius of the pulley (the distance of a point at the edge from the centre)

Substituting into the equation, we find

come closer and become far apart and become steam/gas

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