Analyzing the Motion of a Released Yoyo: Calculating Acceleration and Tension

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In summary, a yoyo with a mass of 179 g, inner radius of 2.60 cm, and outer radius of 3.60 cm is released from rest. Its moment of inertia about the axis perpendicular to the plane of the yoyo and passing through the center of mass is 9.70x10^-5 kgm2. The linear acceleration of the yoyo is determined by the equation a = g*[1 - I/(mr²+I)]. The angular acceleration can be found by the equation α = a/r. The weight of the yoyo is given by W = m*g, and the tension in the rope can be calculated using T = g*I*m/(mr²+I). If a
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Homework Statement


A yoyo with a mass of m = 179 g is released from rest as shown in the figure.

https://s6.lite.msu.edu/res/msu/nagytibo/Mechanics/Rotation/Graphics/001a.gif"

The inner radius of the yoyo is r = 2.60 cm, and the outer radius is R = 3.60 cm, and the moment of inertia about the axis perpendicular to the plane of the yoyo and passing through the center of mass is ICM = 9.70×10-5 kgm2.
1.Determine the linear acceleration of the yoyo.
2. Determine the angular acceleration of the yoyo.
3.What is the weight of the yoyo?
4.What is the tension in the rope?
5.If a 1.04 m long section of the rope unwinds from the yoyo, then what will be the angular speed of the yoyo?

Homework Equations



These are the equations I believe to be correct. However, they are not working out.
1) a = g*[1 - I/(mr²+I)]
2) α = a/r
3) W = m*g
4) T = g*I*m/(mr²+I)
5) w = v/r = √(2ay)/r


The Attempt at a Solution

 
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The figure is not accessible to us. Also, please show how you got the equations that you believe to be correct, then we might be able to point out where you went wrong.
 

Related to Analyzing the Motion of a Released Yoyo: Calculating Acceleration and Tension

1. What is the initial velocity of the yoyo when it is released from rest?

The initial velocity of the yoyo when it is released from rest is zero. This means that the yoyo is not moving at all when it is first released.

2. How does the acceleration of the yoyo change as it falls?

The acceleration of the yoyo increases as it falls due to the force of gravity pulling it downward. This acceleration is constant and equal to 9.8 meters per second squared.

3. What factors affect the motion of the yoyo as it falls?

The motion of the yoyo as it falls is affected by its weight, the force of gravity, and air resistance. Heavier yoyos will fall faster, while air resistance will slow down its descent.

4. What happens to the potential and kinetic energy of the yoyo as it falls?

As the yoyo falls, its potential energy decreases while its kinetic energy increases. This is because potential energy is stored energy, while kinetic energy is energy of motion.

5. What is the final velocity of the yoyo when it hits the ground?

The final velocity of the yoyo when it hits the ground depends on the height from which it was dropped and the acceleration due to gravity. Using the equation v^2 = u^2 + 2as, we can calculate the final velocity, where u is the initial velocity (zero) and s is the distance fallen.

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