Respuesta :
Answer
given,
mass of the solid = 2.50 Kg
speed of translating disk = 3.75 m/s
radius of the disk = 0.1 m
a) transnational Kinetic energy
[tex]KE = \dfrac{1}{2}mv^2[/tex]
[tex]KE = \dfrac{1}{2}\times 2.5 \times 3.75^2[/tex]
[tex]KE =17.58\ J[/tex]
b) rotational kinetic energy.
[tex]KE = \dfrac{1}{2}I\omega^2[/tex]
for disk
[tex]I = \dfrac{1}{2}mr^2[/tex] and v = rω
[tex]KE = \dfrac{1}{2}( \dfrac{1}{2}mr^2)(\dfrac{v}{r})^2[/tex]
[tex]KE = \dfrac{1}{4}mv^2[/tex]
[tex]KE = \dfrac{1}{4}\times 2.5 \times 3.75^2[/tex]
[tex]KE =8.79\ J[/tex]
a) The translational kinetic energy of the disk across the level surface is 17.58J.
b) The rotational kinetic energy of the disk is 8.79J.
Given the data in the question;
- Mass of disk; [tex]m = 2.50kg[/tex]
- velocity; [tex]v = 3.75m/s[/tex]
- radius of disk; [tex]r = 0.100m[/tex]
a)
Translational kinetic energy.
Translational kinetic energy of an object is the work needed to accelerate the object from rest to a given velocity. It is expressed as:
[tex]Translational\ K_E = \frac{1}{2}mv^2[/tex]
We substitute our given values into the equation
[tex]Translational\ K_E = \frac{1}{2}*2.50kg\ *\ (3.75m/s)^2\\\\Translational\ K_E = \frac{1}{2}*2.50kg\ *\ 14.0625m^2/s^2\\\\Translational\ K_E = 17.58kg.m^2/s^2\\\\Translational\ K_E = 17.58J[/tex]
Therefore, the translational kinetic energy of the disk across the level surface is 17.58J
b)
Rotational kinetic energy
Rotational kinetic energy is the energy of rotation of a rotating rigid object or system of particles. its is expressed:
[tex]Rotational\ K_E = \frac{1}{2} Iw^2[/tex]
Where is moment of inertia around the axis of rotation and ω is the angular velocity.
Also, [tex]Moment\ of\ Inertia\ I = \frac{1}{2}mr^2[/tex]
Angular velocity ω is analogous to linear velocity v
So, [tex]v = wr \ and\ w = \frac{v}{r}[/tex]
Hence;
[tex]Rotational\ K_E = \frac{1}{2} * \frac{1}{2}mr^2* (\frac{v}{r})^2\\\\Rotational\ K_E = \frac{1}{2} * \frac{1}{2}mr^2* \frac{v^2}{r^2}\\\\Rotational\ K_E = \frac{1}{2} * \frac{1}{2}mv^2[/tex]
We substitute in our values
[tex]Rotational\ K_E = \frac{1}{2} * \frac{1}{2}*2.50kg * (3.75m/s)^2\\\\Rotational\ K_E = \frac{1}{2} * \frac{1}{2}*2.50kg * 14.0625m^2/s^2\\\\Rotational\ K_E = 8.79kg.m^2/s^2\\\\Rotational\ K_E = 8.79J[/tex]
Therefore, the rotational kinetic energy of the disk is 8.79J
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