Question

A uniform, solid, 100-kg cylinder with a diameter of 1.0 m is mounted so it is free to rotate about fixed, horizontal, frictionless axis that passes through the centers of its circular ends. A 10-kg block is hung from a very light thin cord wrapped around the cylinder's circumference. When the block is released, the cord unwinds and the block accelerates downward
Required:
a. What is the acceleration of the block?
b. Calculate the tension in the string.
c. If the distance between the block and the ground is 3.2 meter, how long does it take to reach the ground?

Answer 1

Answer:

Explanation:

The cylinder effective mass is one half of its actual mass

F = ma

10(9.8) = (10 + 100/2)a

a = 1.633333... ≈ 1.6 N

T = m(g - a)

T = 10(9.8 - 1.6) = 8.2 N

y = ½at²

3.2 = ½(1.6)t²

t = 2.0 s

If one wants to go from basics, The FBD on the cylinder is

τ = Iα

TR = ½MR²α

T = ½MRα

T = ½MR(a/R)

T = ½Ma

The FBD on the block with down being positive is

mg - T = ma

substituting from the cylinder analysis

mg - ½Ma = ma

mg = a(m + ½M)

a = mg / (m + ½M)    which is identical to the equation I used originally.