old-style elevator

2013-07-05 7:41 am
http://session.masteringphysics.com/problemAsset/1260366/3/YF-09-28.jpg

In a charming 19th-century hotel, an old-style elevator is connected to
a counterweight by a cable that passes over a rotating disk 1.50m in
diameter (the figure (Figure 1) ). 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.

1. At how many rpm must the disk turn to raise the elevator at 20.0cm/s ?
unit= rpm

2. To start the elevator moving, it must be accelerated at 18g. What
must be the angular acceleration of the disk, in rad/s^2?

3. Through what angle (in radians ) has the disk turned when it has
raised the elevator 3.20m between floors?

4. Through what angle (in degrees ) has the disk turned when it has raised
the elevator 3.20m between floors?

回答 (1)

2013-07-05 8:21 am
✔ 最佳答案
1. One revolution of the disk = (1.5 x pi) m = 4.712 m
Hence, rotational speed of disk = (0.2/4.712) rev/s = (0.2/4.712) x 60 rpm
= 2.547 rpm

2. 18g = 18 x 9.81 m/s^2 = 176.6 m/s^2
[This is quite a large acceleration. Isn't the given figure correct?]

Angular acceleration = 176.6/0.75 rad/s^2 = 235.4 rad/s^2
[where 0.75 m is the disk radius]

3. Angle at which the disk turned = 3.2/0.75 radians = 4.267 radians

4. Angle turned = 4.267 x 360/(2.pi) degrees = 244.5 degrees


收錄日期: 2021-04-15 15:42:34
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130704000051KK00382

檢視 Wayback Machine 備份