A hoop of radius 0.50 m and a mass of 0.20 kg is released from rest and allowed to roll down an inclined plane. How fast is it moving after dropping a vertical distance of 3.0 m ?
It is 5.4 m/s
V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*3 = 58.8
V = 7.67 m/s.
To find the final velocity of the hoop after dropping a vertical distance of 3.0 m, we can use the conservation of energy principle, which states that the total mechanical energy of an object remains constant if no external forces are acting on it.
The initial potential energy of the hoop at the top of the incline is given by the formula:
PE_initial = m * g * h
Where PE_initial is the initial potential energy, m is the mass of the hoop, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the vertical distance (3.0 m).
PE_initial = 0.20 kg * 9.8 m/s^2 * 3.0 m = 5.88 J
Since the hoop is rolling down the inclined plane, its total mechanical energy is a combination of translational kinetic energy and rotational kinetic energy.
The translational kinetic energy is given by:
KE_translational = (1/2) * m * v^2
Where KE_translational is the translational kinetic energy and v is the velocity of the hoop.
The rotational kinetic energy is given by:
KE_rotational = (1/2) * I * w^2 = (1/2) * (m * r^2) * (v/r)^2
Where KE_rotational is the rotational kinetic energy, I is the moment of inertia of the hoop (given by the formula I = m * r^2), and w is the angular velocity of the hoop.
Since the hoop is rolling without slipping, the velocity of the hoop is related to the angular velocity by the formula v = r * w.
From the conservation of energy principle, we have:
PE_initial = KE_translational + KE_rotational
Substituting in the expressions for translational and rotational kinetic energy, we get:
5.88 J = (1/2) * m * v^2 + (1/2) * (m * r^2) * (v/r)^2
Simplifying the equation and substituting the known values, we get:
5.88 J = (1/2) * 0.20 kg * v^2 + (1/2) * (0.20 kg * (0.50 m)^2) * (v/(0.50 m))^2
Simplifying further, we get:
5.88 J = (1/2) * 0.20 kg * v^2 + 0.050 kg * v^2
5.88 J = 0.10 kg * v^2 + 0.050 kg * v^2
5.88 J = 0.15 kg * v^2
Dividing both sides of the equation by 0.15 kg, we get:
v^2 = (5.88 J) / (0.15 kg) = 39.20 m^2/s^2
Taking the square root of both sides of the equation, we get:
v = √(39.20 m^2/s^2) = 6.26 m/s
Therefore, the hoop is moving with a speed of 6.26 m/s after dropping a vertical distance of 3.0 m.
To find the speed of the hoop after dropping a vertical distance of 3.0 m, we can use the principle of conservation of energy.
The potential energy (PE) of an object at a certain height is converted into kinetic energy (KE) as it falls. The equation for potential energy is:
PE = mgh
Where m is the mass of the object, g is the acceleration due to gravity, and h is the height.
In this case, the potential energy is converted into both rotational and translational kinetic energy as the hoop rolls down the inclined plane.
The equation for the total energy of a rolling hoop is:
E = 1/2Iω^2 + 1/2mv^2
Where I is the moment of inertia of the hoop, ω is the angular velocity, m is the mass, and v is the linear velocity.
For a hoop rolling without slipping, the moment of inertia (I) is given by:
I = mr^2
Where r is the radius of the hoop.
We can rearrange the equation for the total energy to solve for the velocity (v):
v = √((2E - Iω^2)/m)
To find ω, we need to relate it to the tangential velocity (v):
v = rω
Now, let's plug in the given values:
Radius of the hoop (r) = 0.50 m
Mass of the hoop (m) = 0.20 kg
Height (h) = 3.0 m
Acceleration due to gravity (g) = 9.8 m/s^2
First, calculate the potential energy (PE):
PE = mgh
= 0.20 kg * 9.8 m/s^2 * 3.0 m
= 5.88 J
Next, calculate the total energy (E):
E = PE
= 5.88 J
Now, calculate the moment of inertia (I):
I = mr^2
= 0.20 kg * (0.50 m)^2
= 0.05 kg·m^2
Next, calculate the angular velocity (ω) using the tangential velocity equation:
v = rω
ω = v/r
Substituting the expressions for v and r:
v = √((2E - Iω^2)/m)
0.50ω = v
ω = v/0.50
Now, we can substitute ω in the equation for total energy:
v = √((2E - I(v/0.50)^2)/m)
Simplifying the equation:
v = √((2(5.88) - 0.05(v/0.50)^2)/0.20)
This equation relates the velocity (v) to the speed of the hoop after dropping a vertical distance of 3.0 m. By solving this equation, we can find the value of v.