a circular hoop of mass and radius spins like a wheel rotating about its centre with moment of inertia. its period it T
(1) find the kinetic energy in terms of the given parameters
2. if such a hoop rolls with its centre moving at velocity v show that it rolls down an inclined plane with half the acceleration that a frictionless sliding block would have
To find the kinetic energy of the spinning hoop, we need to consider the two types of motion it possesses: rotation and translation.
1) Kinetic Energy of Rotation:
The kinetic energy of rotation is given by the formula: K_rot = 1/2 * I * ω^2, where I is the moment of inertia and ω is the angular velocity.
For a circular hoop, the moment of inertia is given by I = m * r^2, where m is the mass of the hoop and r is the radius. The angular velocity, ω, can be calculated using ω = 2π / T, where T is the period.
Therefore, the kinetic energy of rotation for the hoop is: K_rot = 1/2 * m * r^2 * (2π / T)^2.
2) Kinetic Energy of Translation:
The kinetic energy of translation is given by the formula: K_trans = 1/2 * m * v^2, where v is the velocity of the center of mass of the hoop.
If the hoop rolls with its center moving at velocity v, then the relation between linear velocity and angular velocity is given by v = ω * r.
Therefore, ω = v / r. Substituting this value into the formula for kinetic energy of rotation, we have K_rot = 1/2 * m * r^2 * (v / r)^2 = 1/2 * m * v^2.
3) Total Kinetic Energy:
The total kinetic energy of the hoop, K, is the sum of the kinetic energy of rotation and the kinetic energy of translation.
K = K_rot + K_trans
= 1/2 * m * r^2 * (2π / T)^2 + 1/2 * m * v^2
So, the kinetic energy of the hoop in terms of the given parameters (mass, radius, and period) is K = 1/2 * m * r^2 * (2π / T)^2 + 1/2 * m * v^2.
Now, moving on to the second part of the question:
To show that the hoop rolls down an inclined plane with half the acceleration of a frictionless sliding block:
For a block sliding without friction, its acceleration a_slide is given by a_slide = g * sin(θ), where g is the acceleration due to gravity and θ is the angle of the inclined plane.
For the hoop rolling down the same inclined plane, the acceleration a_roll can be calculated using the following relation: a_roll = (1/2) * a_slide.
This is because, when the hoop rolls, it undergoes both translation and rotation, and a portion of the total acceleration is allocated to each type of motion. The ratio between the two accelerations (a_roll and a_slide) is 1/2.
Therefore, the hoop rolls down the inclined plane with half the acceleration of a frictionless sliding block.