Question

You pull straight up on the string of a yo-yo with a force 0.235 N, and while your hand is moving up a distance 0.18 m, the yo-yo moves down a distance 0.70 m. The mass of the yo-yo is 0.025 kg, and it was initially moving downward with speed 0.5 m/s and angular speed 124 rad/s. (a) What is the increase in the translational kinetic energy of the yo-yo? (b) What is the new speed of the yo-yo? (c) What is the increase in the rotational kinetic energy of the yo-yo? (d) The yo-yo is approximately a uniform-density disk of radius 0.02 m. What is the new angular speed of the yo-yo?

Short answer

a) The increase in translational kinetic energy of yo-yo is 0.007 J.

b) The new speed of yo-yo is 0.9 m/s.

c) The increase in rotational kinetic energy of yo-yo is 0.207 J.

d) The new angular speed of the yo-yo is 313.33 rad/s.

Step-by-step solution

Identification of given data

Given data can be listed below,

• Force,$F=0.235N.$
• Hand moving up distance,$d_u=0.18\hspace{0.33em}m$
• Hand moving down distance,$d_d=0.7\hspace{0.33em}m$
• Mass of yo-yo,$m=0.025\hspace{0.33em}kg$
• Initial speed,$v_i=0.5\hspace{0.33em}m/s$
• Initial angular speed,${\omega }_i=124rad/s$

Translational kinetic energy of the yo-yo

Part a)

By using the law of conservation of energy inthe vertical direction, to find the increase in translational kinetic energy.

$\Delta K{E}_{trans}=(mg-F)d_d$

Substituting 0.025 kg for m , 9.8 m/s² for g, 0.7 for $d_d$, and 0.235 for F in above equation

$$\begin{gathered} {\mathrm{\Delta KE}}_{\mathrm{trans}}=\left(\left(0.025\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\left(0.235\mathrm{N}\right)\right)\left(0.7\mathrm{m}\right) \\ =0.007\mathrm{J} \end{gathered}$$

Thus, the increase in translational kinetic energy of yo-yo is 0.007 J.

Evaluating the speed of the yo-yo

Part b)

By using the change in translation kinetic energy is,

$\Delta K{E}_{trans}=\frac{1}{2}m(v_f^2-v_i^2)$

Where, $v_f$is the final speed of the yo-yo.

Substituting 0.007 J for$∆K{E}_{trans}$ , 0.025 kg for m, and 0.5 m/s for $v_i$in the above equation, we get

$$\begin{gathered} \left(0.007\mathrm{J}\right)=\frac{1}{2}\left(0.025\mathrm{kg}\right)\left(\left({\mathrm{v}}_{\mathrm{f}}^2\right)-{\left(0.5\mathrm{m}/\mathrm{s}\right)}^2\right) \\ {\mathrm{v}}_{\mathrm{f}}^2=\frac{2\times 0.007\mathrm{J}}{0.025\mathrm{kg}}+0.5\mathrm{m}/\mathrm{s} \\ {\mathrm{v}}_{\mathrm{f}}^2=0.81 \\ {\mathrm{v}}_{\mathrm{f}}=0.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Thus, the new speed of the yo-yo is 0.9 m/s.

Rotational kinetic energy

Part c)

The change in the rotational kinetic energy is,

$\Delta K{E}_{rot}=F{d}_u+mg{d}_d-\Delta K{E}_{trans}$

Substituting 0.025 kg for m , 9.8 m/s² for g , 0.7 for $d_d$, 0.235 for F , 0.18 m for $d_u$, and 0.007 J for $∆K{E}_{trans}$in above equation

$$\begin{gathered} {\mathrm{\Delta KE}}_{\mathrm{rot}}=\left(0.235\mathrm{N}\right)\left(0.18\mathrm{m}\right)+\left(0.025\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)\left(0.7\mathrm{m}\right)-\left(0.007\mathrm{J}\right) \\ =0.207\mathrm{J} \end{gathered}$$

Thus, the increase in rotational kinetic energy of yo-yo is 0.207 J.

Evaluating the angular speed

Part d)

By using the change in rotation kinetic energy is,

$\Delta K{E}_{rot}=\frac{1}{2}I({\omega }_f^2-{\omega }_i^2)$

Where $l$is the moment of inertia, and is final angular speed of the yo-yo.

$I=\frac{m{r}^2}{2}$

Substituting 0.025 kg for m , and 0.02 m for r in the above equation.

role="math" localid="1657863207225" $$\begin{gathered} I=\frac{(0.025kg)(0.02\hspace{0.33em}m{)}^2}{2} \\ I=5\times {10}^{6}kg.m^2 \end{gathered}$$

Substituting 5x10⁶ kg.m²for$l$ , 0.207 J for$∆K{E}_{rot}$ , and 124 rad/s for ${\omega }_i$in the equation of change in rotation kinetic energy.

$$\begin{gathered} 0.207J=\frac{1}{2}\left(5\times {10}^{-6}kg.m^2\right)\left(\left({\omega }_f^2\right)-{\left(124rad/s\right)}^2\right) \\ {\omega }_f=313.33rad/s \end{gathered}$$

Thus, the new angular speed of the yo-yo is 313.33 rad/s.