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?

[tex]\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[/tex]

The angular speed of the wheel is 2.38732 rpm

Angular acceleration is given by

[tex]\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[/tex]

The angular acceleration of the wheel is 1.22625 rad/s²

Angular displacement is given by

[tex]\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}[/tex]

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

Lanuel Lanuel · Answer 2 · 2021-11-04T22:34:09+01:00

a. The rpm which the disk must turn to raise the elevator at 25.0 cm/s is 2.39 rpm.

b. The angular acceleration of the disk in [tex]rad/s^2[/tex] is 1.225 [tex]rad/s^2[/tex].

c. The angle (in radians) the disk has turned when it has raised the elevator 2.85 m between floors is 2.85 radians.

d. The angle (in degrees) the disk has turned when it has raised the elevator 2.85 m between floors is163.27 degrees.

Given the following data:

Diameter = 2.00 meter

Speed = 25.0 cm/s

Acceleration = [tex]\frac{1}{8} g[/tex]

Distance = 2.85 m

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8 [tex]m/s^2[/tex]

Radius = [tex]\frac{Diameter}{2} = \frac{2}{2} = 1\;meter[/tex]

Conversion:

Speed = 25.0 cm/s to m/s = [tex]\frac{25}{100} = 0.25 \;m/s[/tex]

a. To find how many rpm the disk must turn to raise the elevator at 25.0 cm/s:

Mathematically, angular speed is given by the formula:

[tex]\omega = \frac{V}{r}[/tex]

Where:

[tex]\omega[/tex] is the angular speed of an object.

r is the radius of an object.

V is the speed of an object.

Substituting the given parameters into the formula, we have;

[tex]\omega = \frac{0.25}{1} \\\\\omega = 0.25\;rad/s[/tex]

Converting the value in rad/s to rpm, we have:

[tex]\omega = 0.25 \times \frac{60}{2\pi} \\\\\omega = \frac{15}{2\times 3.142}\\\\\omega = \frac{15}{6.284}[/tex]

Angular speed, [tex]\omega[/tex] = 2.39 rpm

b. To find the angular acceleration of the disk in [tex]rad/s^2[/tex]:

Mathematically, angular acceleration is given by the formula:

[tex]\alpha = \frac{a}{r} \\\\\alpha = \frac{\frac{1}{8} g}{1}\\\\\alpha = \frac{9.8}{8}[/tex]

Angular acceleration, [tex]\alpha[/tex] = 1.225 [tex]rad/s^2[/tex]

c. To find the angle (in radians) the disk has turned when it has raised the elevator 2.85 m between floors:

[tex]\Theta = \frac{S}{r} \\\\\Theta = \frac{2.85}{1}[/tex]

Angle, [tex]\Theta[/tex] = 2.85 rad.

d. To find the angle (in degrees) the disk has turned when it has raised the elevator 2.85 m between floors:

Angle, [tex]\Theta[/tex] = 2.85 rad.

[tex]Angle, \Theta = 2.85 \times \frac{360}{2\pi}\\\\Angle, \Theta = \frac{1026}{2\times 3.142} \\\\Angle, \Theta = \frac{1026}{6.284}[/tex]

Angle, [tex]\Theta[/tex] = 163.27 degrees.

Read more: https://brainly.com/question/14703849