A hoop of mass M = 4 kg and radius R = 0.4 m rolls without slipping down a hill, as shown in the figure. The lack of slipping means that when the center of mass of the hoop has speed v, the tangential speed of the hoop relative to the center of mass is also equal to vCM, since in that case the instantaneous speed is zero for the part of the hoop that is in contact with the ground (v − v = 0). Therefore, the angular speed of the rotating hoop is ù = vCM/R.

(a) The initial speed of the hoop is vi = 2 m/s, and the hill has a height h = 3.7 m. What is the speed vf at the bottom of the hill?

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To find the final speed (vf) of the hoop at the bottom of the hill, we can use the principle of conservation of mechanical energy. The initial mechanical energy (Ei) of the hoop at the top of the hill is equal to the final mechanical energy (Ef) at the bottom of the hill.

The mechanical energy of the hoop is given by the sum of its kinetic energy (KE) and potential energy (PE):

E = KE + PE

At the top of the hill, the hoop only has potential energy, and at the bottom, it only has kinetic energy. Therefore, we can write:

Ei = PEi
Ef = KEf

The potential energy (PE) is given by the formula:

PE = mgh

Where:
m is the mass of the hoop (4 kg)
g is the acceleration due to gravity (9.8 m/s^2)
h is the height of the hill (3.7 m)

Substituting in the values, we find:

PEi = mgh
PEi = (4 kg)(9.8 m/s^2)(3.7 m)
PEi = 144.08 J

The kinetic energy (KE) is given by the formula:

KE = (1/2)mv^2

Where:
m is the mass of the hoop (4 kg)
v is the final speed of the hoop at the bottom of the hill, which we need to find

Using the conservation of mechanical energy, we equate the initial and final energies:

Ei = Ef
PEi = KEf

Substituting the values, we find:

144.08 J = (1/2)(4 kg)v^2
144.08 J = 2 kg v^2

Simplifying, we have:

v^2 = 144.08 J / 2 kg
v^2 = 72.04 m^2/s^2

Taking the square root of both sides, we find:

v = √(72.04 m^2/s^2)
v ≈ 8.49 m/s

Therefore, the speed vf at the bottom of the hill is approximately 8.49 m/s.

To find the speed vf at the bottom of the hill, we can use the principle of conservation of energy.

1. Gravitational Potential Energy (PE):
At the top of the hill, all of the hoop's energy is in the form of gravitational potential energy. When it reaches the bottom of the hill, this potential energy is converted into kinetic energy.
PE = m * g * h
where m is the mass of the hoop, g is the acceleration due to gravity, and h is the height of the hill.

2. Kinetic Energy (KE):
At the bottom of the hill, all of the hoop's energy is in the form of kinetic energy.
KE = (1/2) * m * vf^2
where vf is the final speed of the hoop at the bottom of the hill.

According to the conservation of energy, the initial potential energy is equal to the final kinetic energy.
PE = KE
m * g * h = (1/2) * m * vf^2

Since the mass m cancels out, we can solve for vf:
vf^2 = 2 * g * h
vf = sqrt(2 * g * h)

Substituting the given values:
g = 9.8 m/s^2 (acceleration due to gravity)
h = 3.7 m (height of the hill)

vf = sqrt(2 * 9.8 * 3.7)
vf = sqrt(72.68)
vf ≈ 8.53 m/s

Therefore, the speed vf at the bottom of the hill is approximately 8.53 m/s.