Respuesta :
Explanation:
It is given that,
Rotational inertia of a yo- yo, [tex]I=1020\ g.cm^2=0.000102\ kg.m^2[/tex]
Mass of yo- yo, m = 83.3 g = 0.0833 kg
Radius of axle, r = 3.71 mm = 0.00371 m
Length of the string, l = 112 cm = 1.12 m
(a) The acceleration of the center of mass is given by :
[tex]a=(\dfrac{g}{1+I/mr^2})[/tex]
[tex]a=(\dfrac{9.8\ m/s^2}{1+0.000102\ kg.m^2\times 0.0833 kg\times (0.00371\ m)^2})[/tex]
[tex]a=9.79\ m/s^2[/tex]
(b) Initially, it is at rest rolls down to the end of the string.
u = 0
Distance, l = 1.12 m
Using second equation of motion as :
[tex]l=ut+\dfrac{1}{2}at^2[/tex]
[tex]l=0+\dfrac{1}{2}at^2[/tex]
[tex]t=\sqrt{\dfrac{2l}{a}}[/tex]
[tex]t=\sqrt{\dfrac{2\times 1.12\ m}{9.79\ m/s^2}}[/tex]
[tex]t=0.47\ s[/tex]
(c) Linear speed, [tex]v=a\times t[/tex]
[tex]v=9.79\ m/s^2\times 0.47\ s[/tex]
v = 4.6013 m/s
(d) Translational kinetic energy, [tex]E_k=\dfrac{1}{2}mv^2[/tex]
[tex]E_k=\dfrac{1}{2}\times 0.0833\ kg\times (4.6013)^2[/tex]
[tex]E_k=0.88\ J[/tex]
(e) Rotational kinetic energy, [tex]E=\dfrac{1}{2}I\omega^2[/tex]
Since, [tex]\omega=\dfrac{v}{r}[/tex]
[tex]\omega=\dfrac{4.6013\ m/s}{0.00371\ m}=1240.24\ rad/s[/tex]
[tex]E=\dfrac{1}{2}\times 0.000102\times (1240.24)^2=78.44\ J[/tex]
(f) Angular speed, [tex]\omega=\dfrac{v}{r}[/tex]
[tex]\omega=\dfrac{4.6013\ m/s}{0.00371\ m}=1240.24\ rad/s[/tex]
Hence, this is the required solution.
The magnitude of the linear acceleration of the yo-yo is 9.1 x 10⁻³ m/s².
The time taken to reach the end of the string is 15.68 s.
The linear speed is 0.142 m/s.
The translational kinetic energy is 8.49 x 10⁻⁴ J.
The rotational kinetic energy is 0.075 J.
The angular speed is 3.82 rad/s.
Magnitude of the linear acceleration
The magnitude of the linear acceleration of the yo-yo is calculated as follows;
[tex]a = \frac{g}{1 + I/mr^2} \\\\a = \frac{9.8}{1 \ + \ (0.000102)/(0.0833 \times 0.00371)^2} = 9.1 \times 10^{-3} \ m/s^2[/tex]
Time of motion
The time taken to reach the end of the string is calculated as follows;
[tex]h = v_0t + \frac{1}{2} at^2\\\\l = 0 + \frac{1}{2} at^2\\\\t = \sqrt{\frac{2l}{a} } \\\\t = \sqrt{\frac{2(1.12)}{9.1 \times 10^{-3}} } \\\\t = 15.68 \ s[/tex]
Linear speed
v = at
v = 9.1 x 10⁻³ x 15.68
v = 0.142 m/s
Translational kinetic energy
[tex]E = \frac{1}{2} mv^2\\\\E = \frac{1}{2} \times 0.0833 \times 0.142^2\\\\E = 8.49 \times 10^{-4} \ J[/tex]
Rotational kinetic energy
[tex]E = \frac{1}{2} I \omega^2\\\\E = \frac{1}{2} \times (0.000102) \times (\frac{0.142}{0.00371} )^2\\\\E = 0.075 \ J[/tex]
Angular speed
The angular speed is calculated as follows;
[tex]\omega = \frac{v}{r} = \frac{0.142}{0.0371} = 3.82 \ rad/s[/tex]
Learn more about rotational kinetic energy here: https://brainly.com/question/25803184
