Question

You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo moves down a distance 0.32 m. The mass of the yo-yo is 0.062 kg, and it was initially moving downward with speed 2.9 m/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?

Answer 1

Given Information:

Force = F = 0.35 N

distance of hand = h = 0.16 m

distance of yo-yo = y = 0.32 m

mass of yo-yo = 0.062 kg

initial velocity = vi = 2.9 m/s

Required Information:

a) Translational kinetic energy = ΔTKE = ?

b) Final velocity = vf = ?

c) Rotational kinetic energy = ΔRKE = ?

Answer:

a) Translational kinetic energy = ΔTKE = 0.0413 Joules

b) Final velocity = Vf = 3.12 m/s

c) Rotational kinetic energy = ΔRKE = 0.251 Joules

Explanation:

a) The change in translational kinetic energy of the yo-yo is given by

ΔTKE = (mg - F)h

Where F is the force exerted by the hand on the yo-yo and h is the distance your hand has moved.

ΔTKE = (0.062*9.81 - 0.35)*0.16

ΔTKE = 0.0413 Joules

b) The new speed of the yo-yo can be found using

ΔTKE = ½mVf² - ½mVi²

Re-arranging the equation for the final velocity of the yo-yo,

½mVf² = ΔTKE + ½mVi²

½*0.062*Vf² = 0.0413 + ½*0.062*(2.9)²

0.031*Vf² = 0.0413 + 0.261

Vf² = 0.3023/0.031

Vf² = 9.75

Vf = √9.75

Vf = 3.12 m/s

c) The increase in the rotational kinetic energy of the yo-yo can be found using

ΔTKE + ΔRKE = mgh + Fd

ΔRKE = mgh + Fd + ΔTKE

ΔRKE = 0.062*9.81*0.16 + 0.35*0.32 + 0.0413

ΔRKE = 0.0973 + 0.112 + 0.0413

ΔRKE = 0.251 Joules

Answer 2

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

$\Delta E_k=W=W_g-W_p$

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

$Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J$

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

$\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2$

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

$v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}$

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

$\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J$

hence, the change in Er is about 1.52J times the initial rotational energy