a hoop of radius .5 m & mass .2 kg is released from rest & rolls down an inclined plane. how fast is it moving when it has dropped a vertical distance of 3 m?
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To find the speed of the hoop when it has dropped a vertical distance of 3 m, we can use the principle of conservation of energy.
First, we need to determine the potential energy of the hoop at the starting position and the kinetic energy of the hoop at the final position.
1. Potential Energy (PE):
The potential energy of the hoop at the starting position is given by the formula:
PE = m * g * h
where m is the mass of the hoop (0.2 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the vertical distance (3 m).
So, PE = 0.2 kg * 9.8 m/s^2 * 3 m = 5.88 J
2. Kinetic Energy (KE):
The kinetic energy of the hoop at the final position is given by the formula:
KE = (1/2) * I * ω^2
where I is the moment of inertia of the hoop and ω is the angular velocity of the hoop.
Since the hoop is rolling without slipping, the linear speed (v) can be related to the angular velocity (ω) by the formula:
v = ω * r
where r is the radius of the hoop (0.5 m).
Therefore, ω = v / r.
To find the moment of inertia (I) of the hoop, we can use the equation for the moment of inertia of a hoop about its center axis: I = (1/2) * m * r^2.
3. Moment of Inertia (I):
I = (1/2) * m * r^2 = (1/2) * 0.2 kg * (0.5 m)^2 = 0.025 kg·m^2
Now, we can write the expression for the kinetic energy at the final position in terms of the linear speed (v) instead of the angular velocity (ω):
KE = (1/2) * I * (v / r)^2 = (1/2) * 0.025 kg·m^2 * (v / 0.5 m)^2 = 0.0125 kg·m^2 * v^2
Now, let's equate the potential energy at the starting position (PE) with the kinetic energy at the final position (KE):
PE = KE
5.88 J = 0.0125 kg·m^2 * v^2
Now, we can solve for the velocity (v):
v^2 = 5.88 J / (0.0125 kg·m^2) = 470.4 m^2/s^2
v = √(470.4 m^2/s^2) ≈ 21.7 m/s
Therefore, when the hoop has dropped a vertical distance of 3 m, it is moving at a speed of approximately 21.7 m/s.