Angular acceleration problem for a pulley used to raise an elevator

In summary: Please pay attention and try to follow his hints.Also, please quote the post to which you are responding when you post. It makes the conversation easier to follow.In summary, the conversation discusses calculating the linear acceleration of an elevator and the tangential acceleration of a point on a rotating disk. The linear acceleration is found by multiplying 1/8 g (1.22625 m/s^2) by the radius (1.25 m), giving a result in units of m^3/s^2, which is incorrect. To find the tangential acceleration, the correct equation is a = rω^2, where ω is the angular velocity. Omega (ω) is not the same as the linear acceleration (a) and should
  • #1
leggythegoose
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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!
 
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  • #2
leggythegoose said:
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.

leggythegoose said:
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?
 
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  • #3
TSny said:
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?
 
  • #4
leggythegoose said:
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"
 
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  • #5
TSny said:
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?
 
  • #7
leggythegoose said:
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.
 
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  • #8
leggythegoose said:
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 ]

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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?
 
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  • #9
leggythegoose said:
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.
 
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FAQ: Angular acceleration problem for a pulley used to raise an elevator

What is angular acceleration in the context of a pulley system?

Angular acceleration refers to the rate of change of angular velocity of the pulley. In the context of a pulley system used to raise an elevator, it describes how quickly the pulley is speeding up or slowing down its rotation.

How do you calculate the angular acceleration of a pulley?

The angular acceleration (α) of a pulley can be calculated using the formula α = τ / I, where τ is the net torque applied to the pulley and I is the moment of inertia of the pulley. The net torque can be derived from the forces acting on the elevator and the radius of the pulley.

What role does the moment of inertia play in determining angular acceleration?

The moment of inertia (I) is a measure of how much resistance a pulley has to changes in its rotational motion. A larger moment of inertia means the pulley will accelerate more slowly for a given torque. It is a crucial factor in determining the angular acceleration of the pulley.

How does the mass of the elevator affect the angular acceleration of the pulley?

The mass of the elevator affects the tension in the cable, which in turn affects the torque applied to the pulley. A heavier elevator increases the tension in the cable, resulting in a greater torque and potentially a higher angular acceleration, depending on the moment of inertia of the pulley.

What is the relationship between linear acceleration of the elevator and angular acceleration of the pulley?

The linear acceleration (a) of the elevator is directly related to the angular acceleration (α) of the pulley by the radius (r) of the pulley. The relationship is given by the equation a = α * r. This means that the linear acceleration of the elevator can be found by multiplying the angular acceleration of the pulley by the radius of the pulley.

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