TORQUE, ROTATION AND TRANSLATION
A Yo-Yo of mass m has an axle of radius b and a spool of radius R . Its moment of inertia about the center can be taken to be I=(1/2)mR2 and the thickness of the string can be neglected. The Yo-Yo is released from rest. You will need to assume that the center of mass of the Yo-Yo descends vertically, and that the string is vertical as it unwinds.
Find the magnitude of the average tension in the string over the course of the Yo-Yo reversing its direction at the bottom of its descent (see figure below).
Express your answer in terms of m, b, R, l and acceleration due to gravity g (enter m for m, b for b, R for R, g for g and pi for π).
http://ocw.mit.edu/courses/physics/8-01t-physics-i-fall-2004/exams/exam03_pract_sol.pdf
thanks Elena you're the best ;)
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yes the answer is 1.47
To find the magnitude of the average tension in the string as the Yo-Yo reverses its direction at the bottom of its descent, we can use the principles of torque, rotation, and translation.
First, let's consider the forces acting on the Yo-Yo at the bottom of its descent. There are two main forces: the tension in the string and the gravitational force.
1. Tension in the String: The tension in the string acts at a distance b from the center of the Yo-Yo, creating a torque. We'll denote this tension as T.
2. Gravitational Force: The gravitational force acts on the Yo-Yo's center of mass, which is assumed to descend vertically. We'll denote the mass of the Yo-Yo as m and acceleration due to gravity as g.
Now, let's analyze the rotational and translational motion of the Yo-Yo. At the bottom of the descent, the Yo-Yo reverses its direction, which means it comes to rest momentarily. This occurs when the torque due to tension equals the torque due to gravity.
The Torque due to Tension:
The torque is defined as the product of force and the perpendicular distance of the force from the rotation axis. In this case, the torque due to tension can be calculated as follows:
τ_tension = T * b
The Torque due to Gravity:
The torque due to gravity tries to rotate the Yo-Yo in the opposite direction. Since the Yo-Yo is at rest at the bottom of its descent, the torque due to gravity is also given by:
τ_gravity = I * α
Where I is the moment of inertia of the Yo-Yo about its center, given as I = (1/2) * m * R^2, and α is the angular acceleration.
At the moment the Yo-Yo comes to a stop, the angular acceleration α is zero. Therefore, τ_gravity = 0.
Setting the torques equal to each other, we have:
T * b = 0 (since τ_tension = τ_gravity)
Solving for T, we find that T = 0.
Hence, the magnitude of the average tension in the string over the course of the Yo-Yo reversing its direction at the bottom of its descent is zero.
This result means that the tension in the string at the bottom of the Yo-Yo's descent is completely balanced by the gravitational force acting on the Yo-Yo's center of mass, resulting in no net tension in the string.