What Linear Speed Must a Hula Hoop Have to Achieve a Specific Kinetic Energy?

What linear speed must a 0.0508-kg hula hoop have if its total kinetic energy is to be 0.190 J? Assume the hoop rolls on the ground without slipping.

To achieve a total kinetic energy of 0.190 J, the hula hoop must have a linear speed of approximately 1.54 m/s. Let's dive deeper into the calculation process to understand how this speed is determined.

Understanding Kinetic Energy and Linear Speed

Kinetic Energy Components: The total kinetic energy (KE) of the hula hoop consists of two components: rotational kinetic energy (KE_rot) and translational kinetic energy (KE_trans). For a rolling hoop without slipping, these two energies are related by the equation: KE = KE_rot + KE_trans Calculating Rotational Kinetic Energy: The rotational kinetic energy of the hoop can be calculated using the formula: KE_rot = (1/2) Iω^2 where I is the moment of inertia and ω is the angular velocity. For a hula hoop, the moment of inertia is given by: I = (1/2) MR^2 where M is the mass of the hoop and R is the radius. Since the hoop is rolling without slipping, the linear speed (v) is related to the angular velocity by the equation: v = ωR Substituting these expressions into the equation for rotational kinetic energy, we get: KE_rot = (1/4) Mv^2 Calculating Translational Kinetic Energy: The translational kinetic energy is given by: KE_trans = (1/2) Mv^2 Equating Total Kinetic Energy: Since the total kinetic energy is the sum of rotational and translational kinetic energies, we can write: 0.190 J = (1/4) Mv^2 + (1/2) Mv^2 Combining the terms, we have: 0.190 J = (3/4) Mv^2 Determining Linear Speed: To find the linear speed (v), we rearrange the equation: v^2 = (4/3) * 0.190 J / M Substitute the values, with M = 0.0508 kg, we can calculate: v^2 = (4/3) * 0.190 J / 0.0508 kg v^2 = 2.360 J/kg v = √(2.360 J/kg) Calculating the square root, we find: v ≈ 1.54 m/s Therefore, the hula hoop must have a linear speed of approximately 1.54 m/s to achieve a total kinetic energy of 0.190 J. This calculation demonstrates the interplay between kinetic energy components and linear speed in the motion of a rolling hula hoop.
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