Torque and energy conservation of a yoyo

In summary, a giant yo-yo with a mass of 480 kg and a radius of 1.9 m was dropped from a height of 57 m in 1993. To find the velocity of descent at the end of the fall, the equation .5mv2+.5Iw2=mgh can be used, but the angular velocity (w) is needed. One approach is to use the force of gravity to calculate the angular acceleration and then use the distance traveled and angular velocity to find the velocity. Another approach is to use the relationship v = R\omega and plug it into the energy conservation equation.
  • #1
scavok
26
0
In 1993, a giant yo-yo of mass 480 kg and measuring about 1.9 m in radius was dropped from a crane 57 m high. Assuming that the axle of the yo-yo had a radius of r=0.1 m, find the velocity of the descent v at the end of the fall.

I know that .5mv2+.5Iw2=mgh, but I don't have a clue how to find w (the angular velocity).

I really don't even know where to start. The only thing I can think of is the force due to gravity creating a torque, which would allow me to solve for the angular acceleration, but even then I would need the amount of time it takes to travel 57m in order to get the angular velocity.

If someone could point me in the right direction I would appreciate it.
 
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  • #2
It is easy to show that [tex]\frac{dz}{dt}=v=R\frac{d\theta}{dt}=R\omega[/tex].
(To show this, simply draw a diagram - If the yo-yo rotates through an angle [tex]d\theta[/tex] what is the consequent change in z?)
 
Last edited:
  • #3
Wouldn't I need the angular velocity, velocity, or time to do anything with that?
 
  • #4
scavok said:
Wouldn't I need the angular velocity, velocity, or time to do anything with that?

According to my last post, [tex] v = R\omega[/tex]. This relation doesn't involve time. Try putting this into your energy conservation equation.
 

Related to Torque and energy conservation of a yoyo

1. What is torque and how does it relate to a yoyo?

Torque is a measure of the force that causes an object to rotate around an axis. In the case of a yoyo, it is the force that is applied to the string to make the yoyo spin.

2. How is energy conserved in a yoyo?

When a yoyo is thrown, it has potential energy due to its height above the ground. As it falls, this potential energy is converted into kinetic energy, causing the yoyo to spin. When the yoyo reaches the end of the string and begins to wind back up, the kinetic energy is converted back into potential energy.

3. Can the energy conservation of a yoyo be affected by friction?

Yes, friction can affect the energy conservation of a yoyo. Friction acts against the motion of the yoyo, causing some of the energy to be converted into heat and sound rather than remaining as potential or kinetic energy.

4. Is there a maximum speed that a yoyo can reach due to energy conservation?

Yes, there is a maximum speed that a yoyo can reach due to energy conservation. This is known as the terminal velocity, which is the maximum speed at which the yoyo can spin while still maintaining a constant energy level.

5. How does the length of the string affect the torque and energy conservation of a yoyo?

The length of the string affects the torque by increasing or decreasing the lever arm, which is the distance between the axis of rotation and the point where the force is applied. A longer string will increase the lever arm and therefore increase the torque. The length of the string also affects the energy conservation by changing the height from which the yoyo is released, thus altering the potential energy and the resulting speed and kinetic energy of the yoyo.

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