The velocity of the 7.7-kg cylinder is 0.49 m/s at a certain instant. What is its speed v after dropping an additional 1.28 m? The mass of the grooved drum is m = 10.5 kg, its centroidal radius of gyration is = 300 mm, and the radius of its groove is ri = 275 mm. The frictional moment at O is a constant 18.4 N·m.

the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

Explanation:

Using the work energy system

The initial kinetic energy is ;

where;

= mass of the cylinder = 7.7 kg

initial velocity of the cylinder = 0.49 m/s

= moment of inertia of the drum about O

mass of the drum = 10.5 kg

radius of gyration = 0.3 m

= angular velocity of the drum

The final kinetic energy is also calculated as:

Similarly, The workdone by all the forces on the cylinder can be expressed as:

where;

g = acceleration due to gravity

h = drop in height of the cylinder

M = frictional moment at O

Finally, using the work energy application;

2.426 + 11.04 = 10.10

13.466 = 10.10

=

= 1.333

Thus, the velocity of the cylinder after dropping an additional 1.28 m is 1.15 m/s

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