A hoop of mass M = 3 kg and radius R = 0.5 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.

Question

A hoop of mass M = 3 kg and radius R = 0.5 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.

(a) The initial speed of the hoop is vi = 3 m/s, and the hill has a height h = 3.8 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 = 3 kg and radius R = 0.5 m, and whose hub has mass m = 1.2 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)).

Answer 1

(a) Well, if the hoop is rolling without slipping down the hill, it's quite the balancing act! But let's get to the numbers. We can use conservation of energy to solve this. The initial kinetic energy of the hoop is given by 1/2 * M * vi^2 (since it's rolling, the rotational kinetic energy is 1/2 * I * ω^2, and for a hoop, I = MR^2).

The potential energy at the top of the hill is M * g * h (where g is the acceleration due to gravity). At the bottom of the hill, the hoop has both translational and rotational kinetic energy, so we can write 1/2 * M * vf^2 + 1/2 * M * R^2 * ω^2 = M * g * h.

Since we know ω = vCM/R, we can replace it in the equation and solve for vf.

(b) Now, let's imagine a bicycle wheel in the same situation! This time, we have to consider the mass of the hub as well. The moment of inertia for a wheel with a hub is given by I = MR^2 + mh^2, where h is the distance from the center of mass to the hub.

Using the same conservation of energy principle, we can replace I in the equation and find the corresponding vf.

But remember, the Clown Bot recommends always wearing a helmet while biking, especially when it comes to solving physics problems involving bicycle wheels! Safety first, my friend!

Answer 2

To solve the problem, we can use the principle of conservation of mechanical energy. At the top of the hill, the hoop/bicycle wheel has gravitational potential energy, which is transformed into kinetic energy at the bottom of the hill.

(a) For the hoop:

The gravitational potential energy at the top of the hill is given by:
PEi = mgh

The kinetic energy at the bottom of the hill is given by:
KEf = (1/2)mvf^2

Since we are assuming no energy losses due to friction or other factors, the total mechanical energy is conserved:
PEi = KEf

Plugging in the given values:
PEi = mgh
KEf = (1/2)mvf^2

Using vi = 3 m/s and h = 3.8 m, and assuming g = 9.8 m/s^2:
PEi = (3 kg)(9.8 m/s^2)(3.8 m)
KEf = (1/2)(3 kg)(vf^2)

Simplifying the expressions:
3.8 mgh = (1/2)(3 kg)(vf^2)

Solving for vf:
vf^2 = (2)(3.8 mgh)/(3 kg)
vf = sqrt((2)(3.8 mgh)/(3 kg))

Now we can calculate vf:

vf = sqrt((2)(3.8 m)(9.8 m/s^2)(3.8 m)/(3 kg))

Calculating vf, we get:

vf ≈ 10.19 m/s

Therefore, the speed vf at the bottom of the hill for the hoop is approximately 10.19 m/s.

(b) For the bicycle wheel:

The gravitational potential energy at the top of the hill is still given by:
PEi = mgh

Since the wheel has rotational motion, we need to consider its rotational kinetic energy as well. The moment of inertia for the bicycle wheel is given by:
I = mR^2 + mh^2

The kinetic energy associated with the rotation of the wheel is given by:
Kerot = (1/2)Iω^2

The total mechanical energy is still conserved:
PEi + Kerot = KEf

Plugging in the given values:
PEi = mgh
KEf = (1/2)mvf^2
I = mR^2 + mh^2
Kerot = (1/2)Iω^2

Using vi = 3 m/s, m = 3 kg, R = 0.5 m, h = 3.8 m, and assuming g = 9.8 m/s^2:
PEi = (3 kg)(9.8 m/s^2)(3.8 m)
KEf = (1/2)(3 kg)(vf^2)
I = (3 kg)(0.5 m)^2 + (3 kg)(3.8 m)^2 = 72.9 kg·m^2
Kerot = (1/2)(72.9 kg·m^2)(ω^2)

Simplifying the expressions:
3.8 mgh = (1/2)(3 kg)(vf^2)
PEi + Kerot = KEf
72.9 kg·m^2 = (1/2)(72.9 kg·m^2)(ω^2)

We can now solve for vf and ω simultaneously. Since we can write ω as vf / R, we can substitute in the value for vf from part (a):

3.8 mgh + (1/2)(72.9 kg·m^2)(vf^2 / R^2) = (1/2)(3 kg)(vf^2)

Substituting vf:

3.8 mgh + (1/2)(72.9 kg·m^2)((10.19 m/s)^2 / (0.5 m)^2) = (1/2)(3 kg)((10.19 m/s)^2)

Calculating vf, we get:

vf ≈ 11.44 m/s

Now, we can solve for ω:

ω = vf / R
ω = 11.44 m/s / 0.5 m

Calculating ω, we get:

ω ≈ 22.88 rad/s

Therefore, the speed vf at the bottom of the hill for the bicycle wheel is approximately 11.44 m/s, and the angular speed ω is approximately 22.88 rad/s.

Answer 3

To solve this problem, we can use the principle of conservation of energy. We can calculate the potential energy at the top of the hill and equate it to the total mechanical energy (kinetic plus potential) at the bottom of the hill.

Let's solve part (a) first.

(a) Initially, the hoop is at rest at the top of the hill, so it has no kinetic energy. Its total mechanical energy only includes potential energy.

The potential energy at the top of the hill is given by:
PE_top = m * g * h

In this case, the mass of the hoop is M, the acceleration due to gravity is g, and the height of the hill is h.

Next, we need to find the final speed (vf) at the bottom of the hill. To do this, we equate the potential energy at the top of the hill to the total mechanical energy at the bottom of the hill:

PE_top = KE_bottom + PE_bottom

At the bottom of the hill, the hoop has both kinetic energy and potential energy. The potential energy at the bottom is zero because the height is zero. The only energy present is kinetic energy.

KE_bottom = (1/2) * I * ω^2

Here, I is the moment of inertia of the hoop with respect to its center of mass, and ω is the angular speed of the hoop (which we can calculate using the given relationship ω = vCM/R, where vCM is the center of mass speed and R is the radius of the hoop).

Since the hoop is rolling without slipping, the moment of inertia can be calculated as I = (1/2) * M * R^2.

Now, let's substitute the values and solve for vf.

PE_top = KE_bottom + PE_bottom
M * g * h = (1/2) * (1/2) * M * R^2 * (vCM/R)^2 + 0

Simplifying the equation:

2 * g * h = (vCM/R)^2

Taking the square root of both sides:

vCM/R = sqrt(2 * g * h)

Multiplying both sides by R:

vCM = R * sqrt(2 * g * h)

Substituting the given values:

vCM = 0.5 m * sqrt(2 * 9.8 m/s^2 * 3.8 m)

Now, we know that the tangential speed of the hoop (v) is equal to the center of mass speed (vCM) because the hoop is rolling without slipping. Therefore, the final speed (vf) at the bottom of the hill is equal to v. Hence,

vf = v = vCM = 0.5 m * sqrt(2 * 9.8 m/s^2 * 3.8 m)

Now, let's solve part (b).

(b) The bicycle wheel consists of a rim and a hub. The rim has mass M = 3 kg and radius R = 0.5 m, and the hub has mass m = 1.2 kg. The spokes have negligible mass.

To find the speed (vf) of the bicycle wheel at the bottom of the hill, we'll still use the principle of conservation of energy.

The potential energy at the top of the hill is the same as before:

PE_top = M * g * h

Now, we need to consider the kinetic energy at the bottom of the hill, but we have to account for both the rim and the hub.

KE_bottom = (1/2) * I_rim * ω^2 + (1/2) * I_hub * ω^2

Since the rim and the hub are rotating with the same angular velocity ω, we can factor it out:

KE_bottom = (1/2) * (I_rim + I_hub) * ω^2

The moment of inertia of the rim with respect to its center of mass is I_rim = (1/2) * M * R^2, as mentioned before.

The moment of inertia of the hub with respect to its center of mass is I_hub = (1/2) * m * R^2, since the hub is rotating about its own center of mass.

Substituting the values:

KE_bottom = (1/2) * ( (1/2) * M * R^2 + (1/2) * m * R^2 ) * ω^2

Since the angular speed ω is the same as before (ω = vCM/R), we can substitute it:

KE_bottom = (1/2) * ( (1/2) * M * R^2 + (1/2) * m * R^2 ) * (vCM/R)^2

Now, equating the potential energy at the top of the hill to the total mechanical energy at the bottom:

PE_top = KE_bottom

M * g * h = (1/2) * ( (1/2) * M * R^2 + (1/2) * m * R^2 ) * (vCM/R)^2

Simplifying the equation:

2 * g * h = ( (1/2) * M * R^2 + (1/2) * m * R^2 ) * (vCM/R)^2

Taking the square root of both sides:

vCM/R = sqrt( (2 * g * h) / ( (1/2) * M * R^2 + (1/2) * m * R^2 ) )

Multiplying both sides by R:

vCM = R * sqrt( (2 * g * h) / ( (1/2) * M * R^2 + (1/2) * m * R^2 ) )

Again, the tangential speed of the wheel (v) is equal to the center of mass speed (vCM) because rolling without slipping is happening. Therefore, the final speed (vf) at the bottom of the hill is equal to v. Hence,

vf = v = vCM = R * sqrt( (2 * g * h) / ( (1/2) * M * R^2 + (1/2) * m * R^2 ) )

Now you can substitute the given values for M, R, m, g, and h to find the speed vf at the bottom of the hill for the bicycle wheel.

To solve the problem, we can use the principle of conservation of mechanical energy. At the top of the hill, the hoop/bicycle wheel has gravitational potential energy, which is transformed into kinetic energy at the bottom of the hill.

To solve this problem, we can use the principle of conservation of energy. We can calculate the potential energy at the top of the hill and equate it to the total mechanical energy (kinetic plus potential) at the bottom of the hill.

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