Solving the Yo-Yo's Limit Angle for Acceleration

AI Thread Summary
The discussion focuses on determining the limit angle θ for a yo-yo's acceleration, where the yo-yo transitions from rolling positively to negatively based on the applied force T at angle θ. The user outlines their approach by incorporating forces such as weight, normal reaction, and static friction into their equations. They express confusion over consistently arriving at nonsensical answers when solving the system of equations, suspecting an error in their sign conventions. The conversation highlights the importance of torque and rotational dynamics in analyzing the yo-yo's motion. Ultimately, the user expresses satisfaction after gaining clarity on the problem.
dany-vai
Messages
6
Reaction score
0

Homework Statement


We have a yo-yo, formed by a little cylinder tied to a little rope, and two big disks as wheels, one for each side of the central cylinder. We do a force T on the rope; its slope is an angle θ over the horizontal x-axis.
We are given the mass m of the cylinder, the mass M of each wheel, the radius of the cylinder R1 and the radius of each wheel R2. We want to know the limit angle θ , for the acceleration to be positive or negative.
I try to explain it better: under a certain angle, the yo-yo have a pure roll in the positive direction, over this angle it goes "back" and rolls towards the negative direction.

The Attempt at a Solution


I put in all the forces, the weight $mg$, the reaction $N$, and the friction force (static, since it's roling) towards the left side.
My equations are:
T*sinθ + N = m*g
T*cosθ - f_{s} = m*a_{cm}
f_{s}*R2 - T*R1 = I_{G}*α = I_{G} * a_{cm} / R2

But I'm always obtaining a non-sense answer solving the system, I guess I'm doing wrong something about the signes. Can someone help me?
Thanks a lot.
 
Physics news on Phys.org


when it rolls, its rotation is around the point where it contacts the ground.
If the torque around this point is zero, it does not roll (alpha = 0).
 


Thank you very much| Now I got it :D
 
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top