Recent content by Megzzy
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Acceleration of a Rolling Cylinder
Sorry for taking so long to get back to you! It's been a busy week. I seem to understand what you have explained so far.- Megzzy
- Post #18
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
radians=degrees x pi/180? Or am I supposed to be making an actual estimate here?- Megzzy
- Post #15
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
I'd get height=sin theta x distance. I don't understand how it plugs into the total energy formula though.- Megzzy
- Post #13
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
all I can think of is sin theta=height/distance- Megzzy
- Post #11
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
alright thank you! I am still unsure of where to go from there and how to connect that with v,w,h and a.- Megzzy
- Post #9
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
So I am picturing the right triangle? d is therefore the hypotenuse and the height is opposite the angle of inclination. I'm sorry this process is taking so long! The question is just really strange to me.- Megzzy
- Post #7
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
I know KE rotational= 1/2Iw^2 KE linear=1/2mv^2 PE=mgh I'm not sure where to go from there.- Megzzy
- Post #5
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
thanks for the help! I have no values so that is why I am having problems I think. I know the moment of inertia of a cylinder is I=1/2MR^2. I'm still lost though.- Megzzy
- Post #3
- Forum: Introductory Physics Homework Help
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Acceleration of a Rolling Cylinder
Homework Statement Prove that a=⅔g sin θ will find the acceleration of a cylinder rolling down an incline of angle θ. Homework Equations a=⅔g sin θ The Attempt at a Solution I don't understand how to do this without numbers. I have a feeling the principle of conservation of mechanical...- Megzzy
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- Acceleration Cylinder Rolling
- Replies: 18
- Forum: Introductory Physics Homework Help
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Why Does the Atwood Machine Equation Deviate from Expected Results?
Oh thank you! That makes much more sense. I'm still really confused about how it relates to the slope and y-intercept though.- Megzzy
- Post #7
- Forum: Introductory Physics Homework Help
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Why Does the Atwood Machine Equation Deviate from Expected Results?
ahhh sorry I misread. R is radians but I don't have radians given.- Megzzy
- Post #5
- Forum: Introductory Physics Homework Help
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Why Does the Atwood Machine Equation Deviate from Expected Results?
R is the radius (2.25 cm).- Megzzy
- Post #3
- Forum: Introductory Physics Homework Help
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Why Does the Atwood Machine Equation Deviate from Expected Results?
Homework Statement I'm currently doing a lab that focuses on a pulley and have got stuck on on part. The goal here is to use the data from the Atwood’s pulley experiment to test the validity of (m1-m2)g=(m1+m2+1/R^2)a. In this equation, take the quantity (m1−m2)g as the y variable and the...- Megzzy
- Thread
- Motion Pulley System
- Replies: 6
- Forum: Introductory Physics Homework Help
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Calculating Impulse Without Mass: A Bouncing Ball Lab Dilemma
The height of the first bounce was 0.799 m. Therefore half that distance is 0.3995 (Where KE and PE are equal). I then took this is used it in the PE equation. The mass of the ball is not changing however so I don't see that it's needed. Therefore PE=mgh PE=0.3995m x 9.8 PE=3.92 Therefore...- Megzzy
- Post #3
- Forum: Introductory Physics Homework Help
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Calculating Impulse Without Mass: A Bouncing Ball Lab Dilemma
Homework Statement I am currently working on a lab that deals with a bouncing ball where I need to find the KE and impulse yet I have no mass given. Homework Equations PE=mgh KE=1/2mv p=mv The Attempt at a Solution I have calculated KE by first finding the PE in the middle of the...- Megzzy
- Thread
- Ball Bouncing ball Momentum
- Replies: 4
- Forum: Introductory Physics Homework Help