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# Yo-Yo Type Problems

## Calculations involving objects that are in free fall but attached to a string.

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Yo-Yo Type Problems

In these types of problems it is very important to realize where the application of the force is acting. Remember that Torque is equal to the Force multiplied by the perpendicular distance from the rotation axis. So if the CM is the rotation axis then there is not torque from gravity, as the distance from the rotation axis is zero. Also, in these types of problems, make sure to use Newton's laws in the linear form (i.e. ΣF=ma\begin{align*} \Sigma F = ma \end{align*}) and also in the rotation form (i.e. Στi=Iα\begin{align*} \Sigma \tau_i = I \alpha\end{align*}). Then you can substitute in for one of the unknowns and solve for the other.

Key Equations

α=τnet/I\begin{align*}\alpha = \tau_{net} / I\end{align*}

angular accelerations are produced by net torques,with inertia opposing acceleration; this is the rotational analog of a=Fnet/m\begin{align*}a = F_{net} / m \end{align*}

τnet=Στi=Iα\begin{align*}\tau_{net} = \Sigma \tau_i = I \alpha\end{align*}

the net torque is the vector sum of all the torques acting on the object. When adding torques it is necessary to subtract CW from CCW torques.

τ=r×F=rF=rF\begin{align*}\tau = r \times F = r _\perp F = rF _\perp\end{align*}

individual torques are determined by multiplying the force applied by the perpendicular component of the moment arm

#### Example

A wooden disk of mass m and radius r has a string of negligible mass is wrapped around it. If the the disk is allowed to fall and the string is held so that it unravels without sliding as the disk is falling, what will be the wooden disk's acceleration?

First, we'll draw a free body diagram for the situation. In the diagram below the force of gravity and the tension from the string have been labeled and are drawn from they points they are acting.

Let's start by applying Newton's second law to this object and we'll let the positive direction be downward.

ΣFmgT=ma=ma\begin{align*} \Sigma F&=ma\\ mg-T&=ma\\ \end{align*}

As you can see, we have two unknowns in this equation(T\begin{align*}T\end{align*} and a\begin{align*}a\end{align*}), so we're going to use Newton's second law for rotation as our second equation and solve for T\begin{align*}T\end{align*}. The rotation axis is at the cm of the cylinder, thus there is no torque from gravity.

ΣτTrT=Iα=Iα=Iαr\begin{align*} \Sigma \tau&=I\alpha\\ Tr&=I\alpha\\ T&=\frac{I\alpha}{r}\\ \end{align*}

Now we will substitute in values for I\begin{align*}I\end{align*} and α\begin{align*}\alpha\end{align*}. In order to calculate the moment of inertia of the wooden disk, we use the known formula for a disk being rotated about it's center.

TTT=12mr2αr=12mr2ar1r=12masubstituting for Isubstituting for α in terms of asimplifying the equation.\begin{align*} T&=\frac{\frac{1}{2}mr^2\alpha}{r} && \text{substituting for } I\\ T&=\frac{1}{2}mr^2\frac{a}{r}\frac{1}{r} && \text{substituting for } \alpha \text{ in terms of } a\\ T&=\frac{1}{2}ma && \text{simplifying the equation.}\\ \end{align*}

We can now plug this back into our original Newton's second law equation from the first step.

mgTmg12maa=ma=ma=23gstart with the equation from we derived in the first stepsubstitute for Tsimplifying and solving for a\begin{align*} mg-T&=ma && \text{start with the equation from we derived in the first step}\\ mg-\frac{1}{2}ma&=ma && \text{substitute for T}\\ a&=\frac{2}{3}g && \text{simplifying and solving for a}\\ \end{align*}

### Review

1. A game of tug-o-war is played … but with a twist (ha!). Each team has its own rope attached to a merry-go-round. One team pulls clockwise, the other counterclockwise. Each pulls at a different point and with a different force, as shown.
1. Who wins?
2. By how much? That is, what is the net torque?
3. Assume that the merry-go-round is weighted down with a large pile of steel plates. It is so massive that it has a moment of inertia of 2000kgm2\begin{align*}2000 \;\mathrm{kg}\cdot m^2\end{align*}. What is its angular acceleration?
4. How long will it take the merry-go-round to spin around once completely?

2. A spherical Yo-Yo is given to you for Christmas. Being the good physicist that you are, you immediately calculate it’s acceleration and then test it with a motion detector. The Yo-Yo, of mass of 200g and a radius of 10cm. is a solid sphere (I = 2/5MR2), with two strings attached, both at a 20° from the top. See figure. The string thickness is negligible compared to the Sphere radius.
1. Calculate the final linear velocity of the Yo-Yo when dropped from a height of 0.5m.
2. Calculate the angular acceleration of the Yo-Yo.
3. If the string thickness is not ignored, how would this change your results in parts a. and b.

1. a. 200N\begin{align*}200 \;\mathrm{N}\end{align*} team b. 40Nm\begin{align*}40 \;\mathrm{Nm}\end{align*} c. 0.02rad/s2\begin{align*}0.02 \;\mathrm{rad/s}^2\end{align*} d. 25s\begin{align*}25 \;\mathrm{s}\end{align*}
2. a. 2.6 m/s b. 72 rad/s2\begin{align*}{rad/s}^2\end{align*} c. Effectively makes r bigger, thus acceleration goes up and thus final velocity goes up.

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