Angular acceleration problem for a pulley used to raise an elevator

[leggythegoose]

Homework Statement

In a charming 19th-century hotel, an old-style elevator is connected to a counterweight by a cable that passes over a rotating disk 2.50 m in diameter. 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. To start the elevator moving, it must be accelerated at 1/8 g. What must be the angular acceleration of the disk, in rad/s^2?

Relevant Equations

constant angular acceleration equations

I tried to multiply 1/8 g (1.22625) by the radius (1.25 m) and got 1.53 rad/s^2. This is actually the linear acceleration of the elevator. How do I get the angular acceleration of the disk? Thanks!

 

[TSny]

I tried to multiply 1/8 g (1.22625) by the radius (1.25 m) and got 1.53 rad/s^2. This is actually the linear acceleration of the elevator.

The linear acceleration of the elevator is 1/8 g.

How do I get the angular acceleration of the disk? Thanks!

What is the tangential acceleration of a point on the rim of the rotating disk? How is the tangential acceleration related to the angular acceleration of the disk?

 

[leggythegoose]

The linear acceleration of the elevator is 1/8 g.What is the tangential acceleration of a point on the rim of the rotating disk? How is the tangential acceleration related to the angular acceleration of the disk?

So I know that r is 1.25 m but how would I find alpha to calculate the tangential acceleration?

 

[TSny]

So I know that r is 1.25 m but how would I find alpha to calculate the tangential acceleration?

You don't need to know $\alpha$ in order to find the tangential acceleration of a point on the rim of the disk. Hint: "the cable does not slip on the disk"

 

[leggythegoose]

You don't need to know $\alpha$ in order to find the tangential acceleration of a point on the rim of the disk. Hint: "the cable does not slip on the disk"

Do I multiply r and omega? So it would be 1.25x1.91 = 2.39 and 2.39 is the tangential acceleration?

 

[Lnewqban]

Do I multiply r and omega? So it would be 1.25x1.91 = 2.39 and 2.39 is the tangential acceleration?

Go back to post #4.

Please, see:
http://hyperphysics.phy-astr.gsu.edu/hbase/rotq.html#rq

Then, try again.

 

[jbriggs444]

I tried to multiply 1/8 g (1.22625) by the radius (1.25 m) and got 1.53 rad/s^2

Note that if you multiply 1.22625 m/s² by 1.25 m you get a result in units of m²/s², not rad/s²

This is a hint that you should not be multiplying those two quantities.

 

[TSny]

Do I multiply r and omega? So it would be 1.25x1.91 = 2.39 and 2.39 is the tangential acceleration?

No. Note that multiplying the units of $\omega$ by the units of $r$ does not give units for acceleration. [EDIT: A similar units problem was noted by @jbriggs444 ]

What is the magnitude of the linear acceleration of point $a$ of the cable? What is the magnitude of the linear (tangential) acceleration of point b on the rim of the disk assuming that the cable does not slip on the disk?

 

[jbriggs444]

Do I multiply r and omega? So it would be 1.25x1.91 = 2.39 and 2.39 is the tangential acceleration?

The number 1.91 which you are calling "omega" -- you got that by multiplying the linear acceleration (a = 1/8 g = 1.22625 m/s²) by the radius (r = 1.25 m) and then again by the radius (r = 1.25 m).

Omega ($\omega$) is the greek letter conventionally used to denote angular velocity, usually with units of rad/s.

The calculation you used cannot possibly yield an angular velocity in rad/s. Instead, it yields a result in units of m³/s².

Then you multiply by 1.25 m yet again, apparently in the hopes that this finally will yield an angular acceleration (in units of m⁴/s²).

Stop multiplying by $r$ already. You are going the wrong way!

@TSny is trying hard to lead you to a correct calculation.