Question

A 1.00 kg solid, uniform sphere rolls without slipping across a level surface, translating at 4.50 m/s. if the sphere's radius is 0.160 m, find the following. hint (a) the sphere's translational kinetic energy (in j) j (b) the sphere's rotational kinetic energy (in j) j

Answer 1

The sphere's translational kinetic energy is 10.125 J and its rotational kinetic energy is also 10.125 J.

Step-by-step explanation:

To find the sphere's translational kinetic energy, we can use the formula K_translational = (1/2) * m * v^2, where m is the mass of the sphere and v is its translational velocity. In this case, m = 1.00 kg and v = 4.50 m/s. Plugging in these values, we get K_translational = (1/2) * 1.00 kg * (4.50 m/s)^2 = 10.125 J.

To find the sphere's rotational kinetic energy, we can use the formula K_rotational = (1/2) * I * w^2, where I is the moment of inertia of the sphere and w is its angular velocity. For a solid sphere rolling without slipping, the moment of inertia is (2/5) * m * r^2, where r is the radius of the sphere. In this case, m = 1.00 kg and r = 0.160 m. Plugging in these values, we get I = (2/5) * 1.00 kg * (0.160 m)^2 = 0.0256 kg·m^2. Since the sphere is rolling without slipping, the angular velocity is related to the translational velocity by the equation w = v/r = 4.50 m/s / 0.160 m = 28.125 rad/s. Plugging in these values, we get K_rotational = (1/2) * 0.0256 kg·m^2 * (28.125 rad/s)^2 = 10.125 J.