Question

In a charming 19th-century hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating disk 2.00 m in diameter (the figure ). The elevator is raised and lowered by turning the disk, and the cable does not slip on the rim of the disk but turns with it.A. At how many rpm must the disk turn to raise the elevator at 25.0 cm/s?B. To start the elevator moving, it must be accelerated at {\textstyle{1 \over 8}}\,g. What must be the angular acceleration of the disk, in {\rm{rad/s}}^2?C.Through what angle (in radians ) has the disk turned when it has raised the elevator 2.85 m between floors?D. Through what angle (in degrees ) has the disk turned when it has raised the elevator 2.85 m between floors?

Answer 1

Answer:

2.38732 rpm

1.22625 rad/s²

163.292°

Explanation:

g = Acceleration due to gravity = 9.81 m/s²

a = Acceleration = $\frac{1}{8}g$

d = Diameter of wheel = 2 m

r = Radius of wheel = $\frac{d}{2}=\frac{2}{2}=1\ m$

v = Speed of elevator = 25 cm/s

Angular speed is given by

$\omega=\frac{v}{r}\\\Rightarrow \omega=\frac{0.25}{1}\\\Rightarrow \omega=0.25\ rad/s=0.25\times \frac{60}{2\pi}\\\Rightarrow \omega=2.38732\ rpm$

The angular speed of the wheel is 2.38732 rpm

Angular acceleration is given by

$\alpha=\frac{a}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}g}{r}\\\Rightarrow \alpha=\frac{\frac{1}{8}\times 9.81}{1}\\\Rightarrow \alpha=1.22625\ rad/s^2$

The angular acceleration of the wheel is 1.22625 rad/s²

Angular displacement is given by

$\theta=\frac{s}{r}\\\Rightarrow \theta=\frac{2.85}{1}\\\Rightarrow \theta=2.85\ rad=2.85\times \frac{360}{2\pi}\\ =163.292\ ^{\circ}$

The angle the disk turned when it has raised the elevator is 163.292°