Question

The cable lifting an elevator is wrapped around a 1.1-m$m$-diameter cylinder that is turned by the elevator's motor. The elevator is moving upward at a speed of 2.3 m/s$\mathrm{m}/\mathrm{s}$ . It then slows to a stop, while the cylinder turns one complete revolution.

How long does it take for the elevator to stop?

Answer 1

Circumfrence of the cylinder:  $C = 2 \pi r$

C = (2)(3.14)(1.1/2) = 3.4558 m

v2 = v02 + 2a(∆x)

In this case the circumfrence length will be ∆x

v = 2.3 m/s

v₀ = 0 

2.3² = 0 + 2a(3.4558)

a = 5.29/(2*3.4558)

a = 0.76539

v = v0 + at

2.3 = 0 + (0.76539)t

t = 2.3/0.732.6

t = 3.0 s

Answer 2

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Answer 3

$IMG_20211116_110023_7~2.jpg$