- #1
Brucezhou
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- 0
Rotational Motion and Torque: Solving for Angular Acceleration
- Thread starter Brucezhou
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Homework Help Overview
The discussion revolves around a problem related to rotational motion and torque, specifically focusing on deriving angular acceleration from torque and moment of inertia. Participants are attempting to prove a relationship involving the vertical component of acceleration for a rod and a ball.
Discussion Character
- Exploratory, Conceptual clarification, Assumption checking
Approaches and Questions Raised
- Participants are discussing the formulation of the torque equation and its simplification. There are questions about the correctness of the initial equations presented, and some participants are seeking clarification on specific variables and their meanings.
Discussion Status
There is an ongoing exploration of the problem with some participants providing guidance on the correct expressions needed for torque and acceleration. Multiple interpretations of the problem setup are being considered, and some participants have acknowledged corrections to earlier approaches.
Contextual Notes
Some participants express uncertainty about the absence of a standard answer key and the implications of starting conditions for the ball and rod. The need for a correct expression for torque is emphasized, and there is mention of specific assumptions regarding the setup of the problem.
- #2
ehild
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How did you get your equation? It is not correct.
You have to prove that the vertical component of the acceleration of the end of the rod is greater than the acceleration of the ball. ehild
You have to prove that the vertical component of the acceleration of the end of the rod is greater than the acceleration of the ball. ehild
- #3
Brucezhou
- 18
- 0
Let Wr cos(sita)=1/3 mr^2* dw/dt.
Simplify it then get:
3g*cos(sita)=r*dw/dt
Simplify it then get:
3g*cos(sita)=r*dw/dt
- #4
Brucezhou
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Could you suggest a way to solving the problem? I don't have a standard answer key and I don't know where I got wrong. I have published my method aboveehild said:
- #5
ehild
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Brucezhou said:
What is W?
ehild
- #6
ehild
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Brucezhou said:
The answer is written in the problem text. You do not need to solve the differential equation. As both the ball and the rod start from rest, you only need to show R what condition is the acceleration of the released ball smaller then the vertical acceleration of the end of the rod.
First you need the correct expression for the torque.
ehild
- #7
Brucezhou
- 18
- 0
weightehild said:
- #8
Brucezhou
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ehild said:
Thank you a lot. I have solved. First my Torque equation is wrong. It should be 1/2 WL cos(sita) since the weight is exerted in the middle. Then it should be
2/3*a(tangent)=g*cos(sita)
2/3*cos(sita)*a(tangent)=g*cos^2(sita)
cos(sita)*a(tangent)=a(vertical to ground)
which should be at least equal g
so
2/3=cos^2(sita)
sita=cos-1(squr(2/3))
- #9
ehild
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It is correct now.
You can improve the formatting to hit the button "go advanced". On the right, you find Greek letters like θ (theta) or mathematical symbols like √ (square root) . Above you see the symbols x2 and x2: you can use x2 to write power (32=9) or inverse function θ=cos-1(2/3). x2 is used when indexing quantities. For example, when you write ∑xi=0. Explore!
ehild
You can improve the formatting to hit the button "go advanced". On the right, you find Greek letters like θ (theta) or mathematical symbols like √ (square root) . Above you see the symbols x2 and x2: you can use x2 to write power (32=9) or inverse function θ=cos-1(2/3). x2 is used when indexing quantities. For example, when you write ∑xi=0. Explore!
ehild