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A solid uniform-density sphere is tied to a rope and moves (without spinning) in a circle with speed 7 m/s. The distance from the center of the circle to the center of the sphere is 1.8 m, the mass of the sphere is 6 kg, and the radius of the sphere is 0.89 m. The angular speed, w=5.62. What is the rotational kinetic energy of the sphere?,

Respuesta :

Kalahira
Speed of the sphere = 7 m/s Distance between center of the circle to the center of the sphere = 1.8 m Mass of the sphere = 6kg Radius of the sphere = 0.89 m Angular speed w = 5.62 Rotational kinetic energy of the sphere = 1/2 I x w^2 I being the moment of inertia, w being the angular speed I = (2/5) x M x R^2 = (2/5) x 6 x 0.89^2 = 1.90 Rotational kinetic energy = 1/2 x 1.9 x 5.62^2 = 30 J
Lanuel

The rotational kinetic energy of the sphere is equal to 37.59 Joules.

Given the following data:

  • Speed = 7 m/s
  • Circular distance = 1.8 m
  • Mass of sphere = 6 kg
  • Radius of sphere = 0.89 m.
  • Angular speed, w = 5.62 rad/s

To find the rotational kinetic energy of the sphere:

First of all, we would determine the moment of inertia (I) of the sphere.

Mathematically, the moment of inertia (I) of an object is given by the formula:

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

Where:

  • I is the moment of inertia (I)
  • m is the mass of an object.
  • r is the radius.

Substituting the given parameters into the formula, we have;

[tex]I = \frac{1}{2} \times 6 \times 0.89^2\\\\I = 3 \times 0.7921[/tex]

Moment of inertia, I = 2.38 [tex]kgm^2[/tex]

Now, we can calculate the rotational kinetic energy of the sphere by using the formula:

[tex]E_R = \frac{1}{2} Iw^2\\\\E_R = \frac{1}{2} \times 2.38 \times 5.62^2\\\\E_R = 1.19 \times 31.58[/tex]

Rotational kinetic energy = 37.59 Joules

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