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 ω = vCM/R.

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?

(b) Replace the hoop with a bicycle wheel whose rim has mass M = 4 kg and radius R = 0.4 m, and whose hub has mass m = 1.3 kg, as shown in the figure. The spokes have negligible mass. What would the bicycle wheel's speed be at the bottom of the hill? (Assume that the wheel has the same initial speed and start at the same height as the hoop in part (a)).

a) The speed vf at the bottom of the hill is given by the equation vf = √2gh, where g is the acceleration due to gravity (9.8 m/s2). Therefore, vf = √2(9.8)(3.7) = 10.2 m/s.

b) The speed of the bicycle wheel at the bottom of the hill is given by the equation vf = √2gh + (m/M)vCM, where m is the mass of the hub, M is the mass of the rim, and vCM is the speed of the center of mass. Therefore, vf = √2(9.8)(3.7) + (1.3/4)2 = 11.2 m/s.

To solve part (a), we can use the principle of conservation of energy. We know that the initial velocity, vi, is 2 m/s and the height of the hill, h, is 3.7 m.

At the top of the hill, the hoop has only potential energy, which can be calculated as:
PEi = M * g * h

At the bottom of the hill, the hoop has both potential and kinetic energy, which can be calculated as:
PEf = 0 (since the hoop is at the bottom)
KEf = (1/2) * M * vf^2 (where vf is the final velocity we need to find)

Since energy is conserved, we can equate the initial and final energies:
PEi = KEf

M * g * h = (1/2) * M * vf^2

Simplifying, we can express vf in terms of g and h:
vf^2 = 2 * g * h
vf = √(2 * g * h)

Substituting the given values, where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is 3.7 m:
vf = √(2 * 9.8 * 3.7)
vf = √(72.04)
vf ≈ 8.49 m/s

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

Now, let's move on to part (b) using the bicycle wheel.

We can still use the principle of conservation of energy, but this time we have to consider both the rotational and translational kinetic energy. The equation for total kinetic energy is modified as follows:

KEf = (1/2) * (M + m) * vf^2 + (1/2) * I * ω^2

Where I is the moment of inertia of the wheel, given by I = MR^2, and ω is the angular velocity.

Since we are assuming that the initial speed and height are the same as in part (a), the potential energy at the top of the hill is still M * g * h.

Equate the initial and final energies:
PEi = KEf

M * g * h = (1/2) * (M + m) * vf^2 + (1/2) * I * ω^2

Substituting the given values, M = 4 kg, m = 1.3 kg, R = 0.4 m, and g = 9.8 m/s^2, we need to find vf. However, we also need to find ω to solve for vf.

We are given that ω = vCM / R, where vCM is the center of mass velocity. Since the wheel is rolling without slipping, vCM = vf as described in the question.

Substituting vCM = vf, we can rewrite the equation as:
M * g * h = (1/2) * (M + m) * vf^2 + (1/2) * I * (vf/R)^2

The moment of inertia can now be written as: I = MR^2 = (4 kg) * (0.4 m)^2 = 0.64 kg·m^2

Now, substitute the values and solve for vf:
M * g * h = (1/2) * (M + m) * vf^2 + (1/2) * (0.64 kg·m^2) * (vf/0.4 m)^2

Simplifying, we can solve for vf:
vf^2 = [(2 * M * g * h) - (0.64 kg·m^2 * (vf/0.4 m)^2)] / (M + m)

This equation is non-linear, so it cannot be easily solved algebraically. However, it can be solved iteratively using numerical methods or graphically.

Therefore, to find the speed vf at the bottom of the hill for the bicycle wheel, more advanced numerical methods or graphical analysis would be needed to find a numerical solution.

To find the speed of the hoop at the bottom of the hill, we can use the conservation of mechanical energy. At the top of the hill, all the initial potential energy is converted into kinetic energy at the bottom of the hill.

(a) For the hoop:
Step 1: Calculate the initial total energy:
E_initial = potential energy + kinetic energy
= mgh + 1/2mv^2
= (4 kg)(9.8 m/s^2)(3.7 m) + (1/2)(4 kg)(2 m/s)^2

Step 2: Calculate the final total energy:
E_final = kinetic energy
= 1/2Iω^2
= 1/2(MR^2)(v_CM/R)^2
= 1/2(4 kg)(0.4 m)^2(v_CM/R)^2

Step 3: Equate the initial and final total energy:
E_initial = E_final
(4 kg)(9.8 m/s^2)(3.7 m) + (1/2)(4 kg)(2 m/s)^2 = 1/2(4 kg)(0.4 m)^2(v_CM/R)^2

Step 4: Solve for v_CM/R, which is equal to ω:
ω = sqrt(((4 kg)(9.8 m/s^2)(3.7 m) + (1/2)(4 kg)(2 m/s)^2) / ((1/2)(4 kg)(0.4 m)^2))

Step 5: Calculate the final linear velocity, vf:
vf = ω * R

Plug in the given values into the equation and solve for vf.

(b) For the bicycle wheel:
The same approach can be applied by modifying the moment of inertia.

Step 1: Calculate the initial total energy:
E_initial = potential energy + kinetic energy
= mgh + 1/2mv^2
= (5.3 kg)(9.8 m/s^2)(3.7 m) + (1/2)(5.3 kg)(2 m/s)^2

Step 2: Calculate the final total energy:
E_final = 1/2Iω^2
= 1/2(MR^2 + mR^2)(v_CM/R)^2

Step 3: Equate the initial and final total energy:
E_initial = E_final
(5.3 kg)(9.8 m/s^2)(3.7 m) + (1/2)(5.3 kg)(2 m/s)^2 = 1/2(MR^2 + mR^2)(v_CM/R)^2

Step 4: Solve for v_CM/R, which is equal to ω:
ω = sqrt(((5.3 kg)(9.8 m/s^2)(3.7 m) + (1/2)(5.3 kg)(2 m/s)^2) / (1/2(MR^2 + mR^2)))

Step 5: Calculate the final linear velocity, vf:
vf = (MR^2 + mR^2)(v_CM/R)

Plug in the given values into the equation and solve for vf.