Question

a cylinder of mass 34.5 kg rolls without slipping on a horizontal surface. At a certain instant, its center of mass has a speed 9.5 m/s. Determine the rotational kinetic energy

Answer 1

Answer:

$\displaystyle 778,40625\:J$

Explanation:

$\displaystyle \frac{1}{2}mv^2 = KE$

* For cylinders, it is unique. Since you have two circular bases, you take half the mass in the formula:

$\displaystyle \frac{1}{4}mv^2 = KE \\ \\ \frac{1}{4}[34,5]9,5^2 = KE → \displaystyle \frac{1}{4}[34,5][90,25] = 778,40625\:J$

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Answer 2

Answer:

778 J

Explanation:

Rotational energy is:

RE = ½ Iω²

For a solid cylinder I = ½ mr².

Rolling without slipping means ω = v/r.

RE = ½ (½ mr²) (v/r)²

RE = ¼ mv²

Plug in values:

RE = ¼ (34.5 kg) (9.5 m/s)²

RE ≈ 778 J

Round as needed.