Two masses of 24 kg and 12 kg are suspended by a pulley that has a radius of 6.2 cm and a mass of 5 kg. The cord has a negligible mass and causes the pulley to rotate without slipping. The pulley rotates without friction. The masses start from rest 4.1 m apart. 4.1 m 6.2 cm 5 kg ω 24 kg 12 kg Determine the speeds of the two masses as they pass each other. Treat the pulley as a uniform disk. The acceleration of gravity is 9.8 m/s 2 . Answer in units of m/s.

since the mass start at rest

Explanation:

Given that :

Two masses of 24 kg and 12 kg are suspended by a pulley

so; let = 24 kg and = 12 kg

radius r = 6.2 cm

mass M = 5.0 kg

The moment of inertia on the pulley is ;

For the force equation with mass m ; we have:

-------- Equation (1)

Force equation of the mass  can be written as:

    -------- Equation (2)

The torque on the pulley is expressed as:

where

Then;

----- Equation (3)

Replacing equation (1) and (2)  into equation (3) ; we have:

235.2 - 72a  = 5a

235.2 = 5a + 72a

235.2 = 77a

a =

a= 3.05 m/s²

Let's consider the mass of the motion, with initial velocity

since the mass start at rest

distance traveled (s) = 2.05 m

acceleration = 3.05 m/s²

Using the formula:

grav force prop 1/r squared 2r all squared is 4rsquared 1/4rsquared quartered

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