- #1
StevenJacobs990
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m1v1 = m2v2
v = rω
m1(rω)1 = m2(rω)2
m1ω1 = m2ω2
Does this make sense?
v = rω
m1(rω)1 = m2(rω)2
m1ω1 = m2ω2
Does this make sense?
Angular Velocity in Linear Momentum Equation
- Thread starter StevenJacobs990
- Start date
AI Thread Summary
The discussion focuses on the relationship between angular velocity and linear momentum, specifically through the equations m1v1 = m2v2 and v = rω. Participants explore the application of these equations in the context of Newton's Cradle, where momentum is transferred through colliding balls. The equations m1(rω)1 = m2(rω)2 and m1ω1 = m2ω2 are presented to illustrate this transfer of momentum. The mechanics of the pendulum swinging down and colliding with the balls are highlighted as a practical example. Overall, the equations effectively describe the dynamics of momentum transfer in this scenario.
The rope is tied into the person (the load of 200 pounds) and the rope goes up from the person to a fixed pulley and back down to his hands. He hauls the rope to suspend himself in the air. What is the mechanical advantage of the system? The person will indeed only have to lift half of his body weight (roughly 100 pounds) because he now lessened the load by that same amount. This APPEARS to be a 2:1 because he can hold himself with half the force, but my question is: is that mechanical...
Some physics textbook writer told me that Newton's first law applies only on bodies that feel no interactions at all. He said that if a body is on rest or moves in constant velocity, there is no external force acting on it. But I have heard another form of the law that says the net force acting on a body must be zero. This means there is interactions involved after all. So which one is correct?
Hello!
I have a question regarding a beam on an inclined plane. I was considering a beam resting on two supports attached to an inclined plane. I was almost sure that the lower support must be more loaded. My imagination about this problem is shown in the picture below.
Here is how I wrote the condition of equilibrium forces:
$$
\begin{cases}
F_{g\parallel}=F_{t1}+F_{t2}, \\
F_{g\perp}=F_{r1}+F_{r2}
\end{cases}.
$$
On the other hand...
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