Falling yo-yo attached to string

In summary, a 410 kg yo-yo with a radius of 1.2 m was dropped from a 63 m high platform, with one end of the string tied to the platform. The yo-yo unwinds as it falls and the axle has a radius of 0.3 m. Using the equations for potential energy and rotational kinetic energy, the velocity of descent at the end of the fall can be found. However, there may be some difficulty in breaking up the rotational energy in this problem.
  • #1
Jeptha
1
0
1.Homework Statement

A giant yo-yo of mass 410 kg measuring about
1.2 m in radius was dropped from a platform
63 m high. One end of the string was tied
to the platform, so the yo-yo unwinds as it
descended. Assuming that the axle of the yo-yo has a
radius of 0.3 m, find the velocity of descent at
the end of the fall. The acceleration of gravity
is 9.81 m/s
Answer in units of m/s.


Homework Equations



M=410 kg
R=1.2 m
h=63 m
r=0.3 m
g=9.81 m/s2

mgh = 0.5Iw2 +0.5mv2

I = 0.5MR2

w = v/R

The Attempt at a Solution



So I've been trying to figure this problem out since last night. Please be patient with me. I've always struggled with problems involving rotational motion for some reason. If I'm understanding this right, the potential energy will be split up into the translational kinetic energy and the rotational kinetic energy. My problem is that I'm having a REALLY hard time visualizing how to break up the rotational energy. I'm assuming that my equation for I (the moment of inertia) is where my problem is because there's torque from gravity and tension from the string, but I don't really understand how I'm supposed to incorporate that.

In my first attempt at a solution, I substituted I and w into my equation for mgh, and after doing some algebra, I came up with v = 2*sqrt(gh/3) and that obviously gave me a wrong answer.

Can someone explain to me what I'm doing wrong conceptually?
 
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  • #2
I really don't understand why my equation for I isn't working. And how should I properly break up the rotational energy in this problem? Thanks
 

1. What causes a yo-yo to fall down when attached to a string?

The force of gravity is what causes a yo-yo to fall down when attached to a string. As the yo-yo is released, the force of gravity pulls it towards the ground.

2. How does the string affect the falling motion of a yo-yo?

The string acts as a guide for the yo-yo, allowing it to move up and down in a controlled motion. Without the string, the yo-yo would simply fall straight to the ground.

3. Why does the yo-yo move back up after it falls down?

When the yo-yo reaches the end of the string, it experiences a change in direction due to the tension in the string. This causes it to move back up towards the hand.

4. Can the length of the string affect the falling motion of a yo-yo?

Yes, the length of the string can affect the falling motion of a yo-yo. A longer string will allow the yo-yo to fall further before reaching the end, while a shorter string will cause it to fall a shorter distance.

5. How does the weight of the yo-yo affect its falling motion?

The weight of the yo-yo can affect its falling motion by changing the force of gravity acting on it. A heavier yo-yo will experience a greater force of gravity and fall faster than a lighter yo-yo.

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