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a yo-yo is constructed of three disks: two outer disks of mass m, radius r and thickness d, and an inner disk of mass m, radius r and thickness d. the yo-yo is suspended from the ceiling and then released with the string vertical. calculate the tension in the string as the yo-yo falls. note that when the center of the yo-yo moves down a distance y, the yo-yo turns through an angle y/r, which in turn means that the angular speed w is equal to vcm/4

User Mrduclaw
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2 Answers

5 votes

Final answer:

The tension in the string as the yo-yo falls can be calculated using the principles of rotational motion. The tension provides the centripetal force needed to keep the yo-yo rotating without slipping. The formula to calculate tension is T = (1/2 * m * r^2) * (a/r).

Step-by-step explanation:

The tension in the string as the yo-yo falls can be calculated using the principles of rotational motion. The tension in the string provides the centripetal force needed to keep the yo-yo rotating without slipping. It can be calculated using the equation T = I * alpha, where T is the tension, I is the moment of inertia, and alpha is the angular acceleration.

In this case, the yo-yo can be considered as a solid cylinder with mass m and radius r. The moment of inertia of a solid cylinder rotating about its axis is given by I = 1/2 * m * r^2. The angular acceleration can be calculated using alpha = a/r, where a is the linear acceleration of the yo-yo.

Therefore, the tension in the string is T = (1/2 * m * r^2) * (a/r) = 1/2 * m * a * r.

User Amit Horakeri
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The tension in the string as the yo-yo falls is
\(T = 3mg + (3)/(2)m\).

The tension in the string can be calculated by considering the forces and motions involved as the yo-yo falls.

First, let's determine the relationship between the linear speed
(\(v_(CM)\)) and the angular speed
(\(\omega\)) of the yo-yo:


\(\omega = (v_(CM))/(r)\)

where r is the radius of the yo-yo.

Given that the center of the yo-yo moves down a distance y and the yo-yo turns through an angle y/r, we can relate linear speed to angular speed.


\(v_(CM) = \omega \cdot r = (y)/(r) \cdot r = y\)

Now, let's consider the forces involved. The tension in the string (T) provides the force necessary to accelerate the yo-yo downward and also generates the torque for rotation.

The net force accelerating the yo-yo downward is the difference between the tension and the gravitational force:


\(T - 3mg = 3ma\)

where:

m = mass of each disk

g = acceleration due to gravity

a = acceleration of the yo-yo's center of mass

The factor of
\(3\) in \(3mg\) accounts for the gravitational force acting on the three disks (two outer disks and the inner disk).

The linear acceleration of the yo-yo's center of mass (a) can be related to the linear speed
(\(v_(CM)\)):


\(a = \frac{{dv_(CM)}}{{dt}} = \frac{{dy}}{{dt}} = v_(CM) \cdot \frac{{dy}}{{dt}} = y \cdot \frac{{dy}}{{dt}}\)

Using the chain rule:
\(\frac{{dy}}{{dt}} = v_(CM)\)

Now, we know that
\(v_(CM) = y\), so:


\(\frac{{dy}}{{dt}} = y\)

This is a differential equation, and to solve it, we'll separate variables and integrate:


\(\int y \, dy = \int dt\)

Integrating both sides:


\(\frac{{y^2}}{2} = t + C\)

Where C is the constant of integration. Considering the initial condition that at t = 0, y = 0 (when the yo-yo is released), we find that C = 0.

So, we get:


\(\frac{{y^2}}{2} = t\)

Now, differentiating both sides with respect to
\(t\) to find \(a = \frac{{dy}}{{dt}}\):


\(y \cdot \frac{{dy}}{{dt}} = 2 \cdot \frac{{dy}}{{dt}} = 1\)

This means
\(a = (1)/(2)\).

Now, going back to our force equation:


\(T - 3mg = 3ma\)

Substitute
\(a = (1)/(2)\) and rearrange:


\(T - 3mg = 3m \cdot (1)/(2)\)


\(T - 3mg = (3)/(2)m\)


\(T = 3mg + (3)/(2)m\)

User Tilish
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