Falling Yo-Yo (angular momentum/energy, torque)

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    Falling Torque
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Homework Help Overview

The problem involves a yo-yo, modeled as a solid cylinder, being dropped from a height of 1.0 m. Participants are discussing the forces and torques acting on the yo-yo to determine the time it takes to hit the ground, focusing on concepts of angular momentum, energy, and torque.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply torque and moment of inertia equations to find the linear acceleration of the yo-yo, leading to a derived acceleration of 2g, which they question. They also explore energy conservation but find it leads to a final velocity instead of time. Other participants question the role of tension in the string and its relationship to gravitational force.

Discussion Status

The discussion is ongoing, with participants raising questions about the assumptions made regarding forces acting on the yo-yo. Some guidance has been offered regarding the importance of considering tension and its effects on the motion of the yo-yo.

Contextual Notes

Participants are navigating the complexities of forces acting on the yo-yo, particularly the relationship between gravitational force and tension, as well as the implications for the motion of the yo-yo as it falls.

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Homework Statement



Today I had a physics midterm and one question that was on it is really bugging me and my friend. We'd really appreciate it if one of the experts here could give us their spin on it (I know...bad pun ;)).

The question is:

A yo-yo (a solid cylinder with a string wrapped around it) is dropped from a height, h=1.0m . How long does it take to hit the ground? Think carefully about all the forces and torques acting on it.

physics-yoyo.png




The Attempt at a Solution



So we know that
\tau=rF=I\alpha

The force is supplied by gravity so F=mg and the moment of inertia of a cylinder is \frac{1}{2}mr^{2} and a = \frac{\alpha}{r} so:

mrg=\frac{1}{2}mr^{2}\frac{a}{r}

Solving for translational (linear) acceleration, I get a=2g which in my mind is incorrect since there's no way that attaching a string to a cylinder could make it accelerate twice as fast as freefall. I didn't know what else to do, so I kept on with that equation and used kinematics to find time of fall:

\Delta s=\frac{1}{2}at^{2}, a=2g
t = \sqrt{2\Delta s a}

Solving with \Delta s=h=1.0m and a=2g I get:

t= \sqrt{4g} = \sqrt{(4)(9.8)} = 6.3s



Does this make any sense? Are we doing it right?


I also tried with energy:

U_{G}=K_{translational}+K_{rotational}
mgh=\frac{1}{2}mv^{2} + \frac{1}{2}I\omega^{2}

but I ended up with only a final velocity (as expected) and not enough information to solve for time.




Somebody please ease our minds.
 
Last edited:
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You probably should have stopped when you got a nonsensical result. The gravitational force on the center of mass of the yoyo is not the only force acting on the yoyo. And it doesn't even produce any torque. What about the tension in the string? That's probably why they said "Think carefully about all the forces and torques acting on it." Try that again.
 
Is the tension not a result of the gravitational force?
 
kayem said:
Is the tension not a result of the gravitational force?

Indirectly, sure. But if the tension were equal to the gravitational force then the forces in the vertical direction would be equal. The yoyo wouldn't fall. It would just hang there. The tension must be less than mg so the yoyo can fall.
 

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