Finding Acceleration in a Constant Angular Acceleration System

Click For Summary

Homework Help Overview

The discussion revolves around finding acceleration in a system with constant angular acceleration, particularly in the context of rotational motion and energy conservation. Participants are exploring the relationships between potential energy, kinetic energy, and rotational dynamics.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Some participants suggest considering conservation of energy as a potential approach, while others express uncertainty about how to apply rotational concepts. There is mention of the relationship between linear and rotational motion, with questions about torque and angular acceleration.

Discussion Status

The discussion is ongoing, with participants sharing different perspectives on how to approach the problem. Some guidance has been offered regarding the use of energy conservation and the need to understand rotational dynamics, but there is no clear consensus on the best method to proceed.

Contextual Notes

Participants note that they have not yet covered all relevant topics in class, such as rotational kinetic energy and torque, which may impact their ability to solve the problem effectively. There is a specific expectation to use a rotational constant acceleration approach for the homework.

jtw2e
Messages
27
Reaction score
0

Homework Statement



5596521604_fe49cccea3_b.jpg


Homework Equations


? constant acceleration equations I guess


The Attempt at a Solution



They didn't get to this part in class today, yet it's expected on our online homework due tonight. No idea how to even get started.
 
Physics news on Phys.org
For these questions you can consider conservation of energy.

PE + KE = constant.
 
rock.freak667 said:
For these questions you can consider conservation of energy.

PE + KE = constant.

jtw2e: Your post title makes my think you might not yet have been introduced to rotational kinetic energy. For an alternate approach: consider the relationship between how fast the block lands and how fast the disk rotates. Do you see that you can work out the torque acting on the disk and find the rotational speed?
 
Fewmet said:
jtw2e: Your post title makes my think you might not yet have been introduced to rotational kinetic energy. For an alternate approach: consider the relationship between how fast the block lands and how fast the disk rotates. Do you see that you can work out the torque acting on the disk and find the rotational speed?

Thank you. As of today we began rotational kinetic energy but did not get very much covered in the topic. We are supposed to begin torque on Friday.
 
rock.freak667 said:
For these questions you can consider conservation of energy.

PE + KE = constant.

While I would like to use CoE, we're supposed to find our answers with rotational constant acceleration approach. I've actually been doing my other homework with CoE just to get the answers turned in. I don't know how to find them with this rotational stuff.
 
If you don't go with the energy approach, you'll want to find the angular acceleration of the pulley system.

You can use these equations:
[tex] \begin{align}<br /> \vec{\tau} &= \vec{r} \times \vec{F} \\<br /> \vec{\tau}_{net} &= I\vec{\alpha}_{net}<br /> \end{align}[/tex]
In these equations, [itex]\vec{\tau}[/itex] is torque, I is moment of inertia, [itex]\vec{\alpha}[/itex] is angular acceleration, and r is the distance vector from the axis of rotation to the force F.

These are the rotational analogues to force and Newton's second law. You can use Eq. (2) on the pulley system to find the net angular acceleration, which will give the system's acceleration. To find the linear acceleration of the block, [itex]\vec{a} = r\vec{\alpha}[/itex].
 

Similar threads

Replies
15
Views
2K
  • · Replies 20 ·
Replies
20
Views
3K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 2 ·
Replies
2
Views
6K
  • · Replies 2 ·
Replies
2
Views
3K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 28 ·
Replies
28
Views
7K
Replies
1
Views
6K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K