Question

A yo-yo has a rotational inertia of 950gcm² and a mass of 120g. Its axle radius is 3.2mm, and its string is 120cm long. The yo-yo rolls from rest down to the end of the string. (a) What is the magnitude of its linear acceleration? (b) How long does it take to reach the end of the string? As it reaches the end of the string, (c) What is its linear speed? (d) What is its translational kinetic energy? (e) What is its rotational kinetic energy? (f) What is its angular speed?

Short answer

1. Magnitude of linear acceleration is 12.5cm/s²
2. Time to reach the end of string is 4.38 s
3. Linear speed at end of string is 54.8 cm/s
4. Translational kinetic energy at end of string is $1.8\times {10}^{-2}J$
5. Rotational kinetic energy at end of string is 1.4 J
6. Angular speed at end of string is$1.7\times {10}^{2}rad/s$

Step-by-step solution

Identification of given data

1. Rotational inertia is $l_{com}=950g.c{m}^2$,
2. Mass is $M=120grams$.
3. The radius of axel is $R_0=3.2mmor0.32cm$.
4. String length $Y=120cm$

To understand the concept

Here, the problem is based on the concept of rotational mechanics and static frictional force due to the rolling of the object along the length of the string. Thus, the force acting on the body is considered to be acting on the center of mass of the system to initiate a linear acceleration of the body that in turn, gives the translational and rotational motion along the string due to its displacement.

Formulae:

The acceleration of the center of mass,

$a_{com}=-\frac{g}{1+\left({\displaystyle \frac{l_{com}}{M{R}_0^2}}\right)}$

where, g is the acceleration due to gravity, l_com is the moment of inertia of the center of mass, R₀ is the mass of the body, M is the radius of the body.

The distance travelled by the body according second law of kinematics,

$y_{com}=v_{0}t+\frac{1}{2}{a}_{com}{t}^2$

Where, v₀ is the initial velocity, t is the time taken, a_com is the acceleration of the center of mass.

The final velocity of the center of mass according to first law of kinematics,

$v_{com}=v_0+a_{com}t$

Where, v₀ is the initial velocity, t is the time taken, a_com is the acceleration of the center of mass.

The angular velocity of the body,

$\omega =\frac{v_{com}}{R_0}$

Where, v_com is the velocity of the center of mass, R₀ is the radius of the circular path.

The translational kinetic energy of the body,

role="math" localid="1661230662609" $K_{trans}=\frac{1}{2}M{v^2}_{com}$

Where, M is the mass of the body, v_com is the velocity of center of mass.

The rotational kinetic energy of the body,

$K_{rot}=\frac{1}{2}{l}_{com}{\omega }^2$

Where, l_com is the moment of inertia of the center of mass, $\omega$is the angular velocity of the body.

(a) Determining the magnitude of linear acceleration

Referring to figure,

$f_s-Mg\sin \theta =Ma$

Consider torque caused by localid="1661230914054" $F_N,F_N\sin \theta and{f}_s$about an axis passing through the center of mass of the sphere.

We have, torque caused by $F_{N}and{F}_g\sin \theta$as zero because these forces are acting at the center of mass of the sphere and so the position vector would be zero.

Therefore,

$\begin{array}{rcl}\tau & =& l_{com}\alpha \\ R{f}_s& =& l_{com}\alpha \\ f_s& =& \frac{l_{com}\alpha }{R}\\ \alpha & =& -\frac{a}{R}\end{array}$

Negative sign arises because of the fact that the acceleration is along the negative x-axis

$f_s=-\frac{l_{com}a}{R^2}$

Therefore,

$\begin{array}{rcl}-\frac{l_{com}a}{R^2}-Mg\sin \theta & =& Ma\\ -Mg\sin \theta & =& Ma+\frac{l_{com}a}{R^2}\\ -Mg\sin \theta & =& a\left(\frac{M{R}^2+l_{com}}{R^2}\right)\\ a& =& \frac{Mg\sin \theta }{M\left(\frac{l_{com}}{M{R}^2}\right)}\end{array}$

So for $\theta ={90}^o$, we have

$\begin{array}{rcl}{a}_{com}& =& -\frac{g}{1+\left({\displaystyle \frac{l_{com}}{M{R}_0^2}}\right)}\\ & =& -\frac{980cm/s^2}{1+\left({\displaystyle \frac{950g-c{m}^2}{\left(120g\right){\left(0.32cm\right)}^2}}\right)}\\ & =& -12.5cm/s^2\end{array}$

Magnitude of acceleration is 12.5cm/s².

(b) Determining the time to reach the end of string

$y_{com}=v_{0}t+\frac{1}{2}{a}_{com}{t}^2$

We take origin at initial position and initial velocity as zero.

$\begin{array}{rcl}120cm& =& 0+\frac{1}{2}\left(12.5cm/s^2\right)t^2\\ t& =& \sqrt{\frac{2\left(120cm\right)}{12.5cm/s^2}}\\ & =& 4.38sec\end{array}$

(c) Determining the linear speed at end of string

$\begin{array}{rcl}{v}_{com}& =& v_0+a_{com}t\\ & =& 0+\left(-12.5cm/s^2\right)\left(4.38s\right)\\ & =& -54.8cm/s\end{array}$

Magnitude of linear velocity is

$v_{com}=54.8cm/s$

(d) Determining the translational kinetic energy

$\begin{array}{rcl}{K}_{trans}& =& \frac{1}{2}\left(0.120kg\right){\left(0.548m/s\right)}^2\\ & =& 1.8\times {10}^{-2}J\end{array}$

(e) Determining the rotational kinetic energy

For this calculation, we need to find the angular velocity, whichis givenas,

$\omega =\frac{V_{com}}{R_0}$

Rotational kinetic energy is given as,

$\begin{array}{rcl}{K}_{rot}& =& \frac{1}{2}{l}_{com}{\omega }^2\\ & =& \frac{1}{2}\left(9.50\times {10}^{-5}kg{m}^2\right)\frac{0.548m/s}{\left(3.2\times {10}^{-3}m\right)}\\ & =& 1.4J\end{array}$

(f) Determining the angular speed

$\begin{array}{rcl}\omega & =& \frac{v_{com}}{R_0}\\ & =& \frac{0.548m/s}{3.2\times {10}^{-3}m}\\ & =& 1.7\times {10}^{2}rad/sec\end{array}$