Respuesta :
Answer:
(a) 147 j (b) 73.5 J (C) 220.5 J
Explanation:
Mass of the cylinder m =6 kg
velocity = 7 m/sec
(a) transnational energy [tex]KE =\frac{1}{2}mv^2=0.5\times 6\times 7^2=147 \ j[/tex]
(b) Rotational kinetic energy is given by [tex]KE =\frac{1}{2}I\omega ^2[/tex] where I is moment of inertia and ω is the angular velocity
Moment of inertia [tex]I=\frac{1}{2}mr^2[/tex]
Putting the value of I in rotational kinetic energy formula
[tex]KE=\frac{1}{2}\frac{1}{2}mr^2\omega ^2[/tex]
[tex]KE=\frac{1}{4}m(rw)^2[/tex]
[tex]KE=\frac{1}{4}m(v)^2[/tex] as v=ωr
So rotational kinetic energy [tex]=\frac{1}{2}\times translational\ kinetic \ energy=0.5\times 147=73.5j[/tex]
(c) Total energy = 147+73.5=220.5 j
a) The translational kinetic energy of the cylinder's center of mass is 147J.
b) The rotational kinetic energy about the cylinder's center of mass is 7.35J
c) The total energy of the cylinder's center of mass is 220.5J.
Given the data in the question;
- Mass of cylinder; [tex]m = 6.0kg[/tex]
- Speed of the center of mass of the cylinder; [tex]v = 7.0m/s[/tex]
a) The translational kinetic energy of the cylinder's center of mass.
Translational kinetic energy is the work needed to accelerate an object or particle from rest to a given velocity. It is expressed as;
[tex]Translational\ K_E = \frac{1}{2}mv^2[/tex]
Where m is the mass of the object and v is its velocity
We substitute of values into the equation
[tex]Translational\ K_E = \frac{1}{2}*6.0kg\ *\ (7.0m/s)^2\\\\Translational\ K_E = \frac{1}{2}*6.0kg\ *\ 49m^2/s^2\\\\Translational\ K_E = 147kg.m^2/s^2\\\\Translational\ K_E = 147J[/tex]
Therefore, the translational kinetic energy of the cylinder's center of mass is 147J.
b) The rotational kinetic energy about the cylinder's center of mass
Rotational kinetic energy is the kinetic energy of rotation of a rotating rigid objector system of particles. its is expressed:
[tex]Rotational K_E = \frac{1}{2}Iw^2[/tex]
Where [tex]I[/tex] is moment of inertia around the axis of rotation and ω is the angular velocity.
We know that;
[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}\ *\ 6.0kg\ *\ (7.0m/s)^2\\\\Rotational K_E = \frac{1}{2}\ *\ \frac{1}{2}\ *\ 6.0kg\ *\ 49m^2/s^2\\\\Rotational K_E = 7.35 kg.m^2/s^2\\\\Rotational K_E = 7.35J[/tex]
Therefore, the rotational kinetic energy about the cylinder's center of mass is 7.35J
c) Total energy
Total energy is the sum of the translational kinetic energy and the he rotational kinetic energy of the cylinder's center of mass.
[tex]T_E = Translational\ K_E + Rotational\ K_E[/tex]
[tex]T_E = 147J + 7.35J\\\\T_E = 220.5 J[/tex]
Therefore, the total energy of the cylinder's center of mass is 220.5J.
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