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?

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 is 1.225 .

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:

We know that the acceleration due to gravity (g) of an object on planet Earth is equal to 9.8

Radius =

Conversion:

Speed = 25.0 cm/s to m/s =

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:

Where:

Substituting the given parameters into the formula, we have;

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

Angular speed, = 2.39 rpm

b. To find the angular acceleration of the disk in :

Mathematically, angular acceleration is given by the formula:

Angular acceleration, = 1.225

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

Angle, = 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, = 2.85 rad.

Angle, = 163.27 degrees.

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see the picture for the vector

the magnitude of the vector by pythagorean theorem

r = √( (-10)^2 + (3)^2 ) = √109

magnitude = √109 units

the direction angle between vector and x-axis is

tan^(-1)(3/10) = 16.699

so the direction is approximately 16.7 degrees north of west