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What is the acceleration a of the center of the hoop?

User AlphaCR
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Final Answer:

The acceleration (a) of the center of the hoop is g/2, where g is the acceleration due to gravity.

Step-by-step explanation:

The acceleration of the center of a hoop rolling without slipping can be determined by considering the rotational and translational motion. For a hoop with radius R and mass m rolling without slipping, the net acceleration a of its center is given by a = g/2, where g is the acceleration due to gravity.

To understand this, consider the sum of translational and rotational kinetic energies. The translational kinetic energy (1/2 * m * v²) is related to the rotational kinetic energy (1/2 * I * ω²) for the hoop, where I is the moment of inertia and ω is the angular velocity. Since the hoop is rolling without slipping, v = Rω, and by substituting this into the kinetic energy equation, you can relate ω to v.

The net acceleration a is then expressed as a = dv/dt = R * dω/dt. By combining the expressions for v and ω, you can derive a = g/2. This result highlights the unique relationship between the linear and angular motions of a hoop rolling without slipping, demonstrating that the acceleration of the center is half of the acceleration due to gravity.

User Deepender Singla
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