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A 10-kg cylinder rolls without slipping on a rough surface. at an instant when its center of gravity has a speed of 10 m/s, determine (a) the translational kinetic energy of its center of gravity, (b) the rotational kinetic energy about its center of gravity, and (c) its total kinetic energy.

Respuesta :

Answer:

Part a)

[tex]K_t = 500 J[/tex]

Part b)

[tex]K_r = 250 J[/tex]

Part c)

[tex]K = 750 J[/tex]

Explanation:

Part a)

As we know that cylinder is rolling without slipping

So we have

[tex]v = R\omega[/tex]

now we have

[tex]v = 10 m/s[/tex]

[tex]m = 10 kg[/tex]

now we know that translational kinetic energy is given as

[tex]K_t = \frac{1}{2}mv^2[/tex]

[tex]K_t = \frac{1}{2}(10)(10^2)[/tex]

[tex]K_t = 500 J[/tex]

Part b)

Now for rotational kinetic energy we know that

[tex]K_r = \frac{1}{2}I \omega^2[/tex]

[tex]K_r = \frac{1}{2}(\frac{1}{2}mR^2)(\frac{v}{R})^2[/tex]

[tex]K_r = \frac{1}{4}mv^2[/tex]

[tex]K_r = \frac{1}{4}(10)(10^2)[/tex]

[tex]K_r = 250 J[/tex]

Part c)

Total kinetic energy is given as

[tex]K = K_r + K_t[/tex]

[tex]K = 500 + 250[/tex]

[tex]K = 750 J[/tex]

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