The idea was that you would transfer a dime from one mass to the other to get an increased acceleration from one trial to the next. It accelerates the system twice the weight on the dime faster as its removed from one and placed on the other, so the mass of the system is always constant. Now...
Homework Statement
I am given the equation (m1 – m2)g = (m1 + m2 + I/R2)a and the experiment is to validate this equation.
Homework Equations
The Attempt at a Solution
After following the lab guide, it tells you to plot the weight difference (m1– m2)g against acceleration and determine...
Right. Just so I am understanding that correctly.
In minutes my uncertainty is +/- 0.2, therefore on 20.0 minutes, it would remain as 20.0 +/- 0.2 min (both have the same amount of decimal places)
In hours the uncertainty would be +/- 0.003, where 20.0 min = 0.333 hr, would be 0.333 +/- 0.003...
Homework Statement
I am starting to confuse myself with the proper use of SF. I am to convert time from minutes to hours, keeping in mind proper SF
Homework Equations
conversion factor: 1 min = 1/60 hr
The Attempt at a Solution
The timing error is +/- 0.2 (1SF) = 0.003hr (1SF)
20.0...
If we know angular acceleration of the first half of the rotation (to horizontal), how could I use this value to figure out the angular velocity at the bottom? I'm assuming angular acceleration would be constant throughout??? I looked at some equations, but nothing that gave me the answer I had...
So I get:
ΣF = mg - FT = ma
FT = m(g - a)
= .3kg (9.8 - 0.12m/s^2)
= 2.9 N
Then for part c (mass of cylinder), I tried using α = a/r, τ = FTr, I = 1/2MR^2, and τR = Iα to solve for mass, but I'm still left with r from α = a/r
FTr x R = I α
FT R^2 = 1/2MR^2 α
FT = 1/2M a/r
OK.
So, am I correct in saying:
y = 1/2at^2
0.54m = 1/2a(3)^2
a = 0.12m/s^2
Then,
ΣF = FT - mg = ma
FT = m(g+a)
= .3kg (9.8 + 0.12m/s^2)
= 2.976 Nm
I just realized with a being so small it didn't make much of a difference. Did I get a wrong?
I am working on this problem, and when solving for the tension in the string, am I assuming it is prior to release? where: ΣF = FT - mg = 0, so T = 2.94 Nm? Or should I be solving for ΣF = FT - mg = ma and use Δy of 0.54m in 3s to solve for a?
I was thinking that the F of 35N would have to be resolved into components and the only one being effective in producing rotational motion would be F sin θ. I'm pretty sure I'm just mixing up the theory behind torque.
Homework Statement
Calculate the net torque about the axle of the wheel shown in https://jigsaw.vitalsource.com/books/9781269392464/content/id/ch08fig39 [Broken]. Assume that a friction torque of 0.40 m · N opposes the motion. Where F=18N (in photo).
The picture is here...
Homework Statement
An airplane makes a circular turn of radius 9 km at a constant speed of 650 km/h. Calculate the magnitude of the plane's a. angular velocity, b. centripetal acceleration, c. angular acceleration, d. tangential acceleration
Homework Equations
w = v/r
aR = w^2r
The Attempt at...
Oh, thank you. It's been a long time since I took any math, which makes application of physics theory at times challenging.
I went back and solved for v2' and v1' using θ = 30 and Φ=60°.
I hope this is ok:
0 = v1' sinθ - v2' sin Φ
0 = v1' sin30 - v2' sin 60
0 = 0.5v1' - 0.866v2'
v2' = 0.5v1' /...
@ehild I'm following everything until "But the term in the parentheses is equal to cos(θ+Φ) =0 so θ+Φ=90°." I understand you're proving what was stated in previous posts, which would make Φ = 60°. However, I have no idea how you went from 2 u1 u2 (cosθ cosΦ-sinθ sinΦ) =0 to cos(θ+Φ) =0.
If v2^2 = v1^2 - 2v1v1'cosθ - v1'^2
and (v1 - v1')(v1 + v1') = v2'^2
Can I equate them and say:
v1^2 - 2v1v1'cosθ - v1'^2 = (v1 - v1')(v1 + v1')
???
I'm quite certain I will chuckle once I figure out the other method.
Right now I am not completely following the CM context - probably because my brain is too tired for new material. However, I tried the math again and it looks ugly, but I am hoping it's on the right track:
(v1 - v1'cosθ)^2 - v1'^2sin^2θ = v2'^2cos^2θ + v2'^2sin^2θ
v1^2 - 2v1v1'cosθ +...