A hoop of radius 1.4 m and mass 11.5 kg has a small ball stuck to its inside surface. The ball has a radius of one quarter that of the hoop, and twice the hoop’s mass. The hoop may roll freely, without slipping. Similarly, the ball may roll along the inside of the hoop without slipping. Suppose the system starts out with the ball on the rightmost side of the hoop horizontally aligned with the hoop’s centre, and that the hoop’s centre is at x=0 along the x-axis. The system is then released, and the ball rolls down to the bottom of the hoop. Calculate the location along the x-axis ,in m, of the ball when it reaches the bottom of the hoop.
The location of the ball when it reaches the bottom of the hoop is x = -1.4 m.
To calculate the location along the x-axis of the ball when it reaches the bottom of the hoop, we need to consider the conservation of angular momentum.
First, let's calculate the initial angular momentum of the system. The angular momentum of the hoop is given by the product of its moment of inertia (I) and its angular velocity (ω). The moment of inertia of a hoop of radius R and mass M rotating about its axis is given by I = MR^2.
Given that the radius of the hoop is 1.4 m and its mass is 11.5 kg, we can calculate its moment of inertia:
I_hoop = M_hoop * R_hoop^2
= 11.5 kg * (1.4 m)^2
= 22.372 kg·m^2
The angular velocity of the hoop can be calculated as the velocity along the circumference divided by the hoop's radius. Since the hoop is rolling freely without slipping, the velocity along the circumference is related to its linear velocity (v_hoop) by the equation v_hoop = ω_hoop * R_hoop.
Considering that the hoop has a radius of 1.4 m and is rolling freely, we can assume that the linear velocity of the hoop is equal to the velocity of the ball when it reaches the bottom of the hoop. So, we can take v_hoop as the unknown and solve for it later.
The angular momentum of the ball is given by the product of its moment of inertia (I_ball) and its angular velocity (ω_ball). The moment of inertia of a solid ball rotating about its axis is given by I = 2/5 * M * R^2, where M is the mass of the ball and R is its radius.
Given that the radius of the ball is one-fourth the radius of the hoop (1.4 m / 4 = 0.35 m) and its mass is twice the mass of the hoop (2 * 11.5 kg = 23 kg), we can calculate its moment of inertia:
I_ball = (2/5) * M_ball * R_ball^2
= (2/5) * 23 kg * (0.35 m)^2
≈ 1.29 kg·m^2
The angular velocity of the ball is related to its linear velocity (v_ball) by the equation v_ball = ω_ball * R_ball. Since the ball is rolling along the inside of the hoop without slipping, the linear velocity of the ball is equal to the linear velocity of the hoop. So, we can consider v_hoop as v_ball.
Now, let's use the conservation of angular momentum to find the velocity of the hoop/ball system when the ball reaches the bottom of the hoop.
The angular momentum of the system is conserved before and after the ball reaches the bottom of the hoop, as there are no external torques acting on the system.
Initial angular momentum (L_initial) = Final angular momentum (L_final)
The initial angular momentum of the system is given by the sum of the angular momentum of the hoop and the angular momentum of the ball.
L_initial = I_hoop * ω_hoop + I_ball * ω_ball
The final angular momentum of the system is the angular momentum of the hoop when the ball reaches the bottom, assuming the hoop is not rolling at that point.
Since the hoop is not rolling at the bottom, its angular velocity (ω_hoop) is zero.
L_final = I_hoop * ω_hoop + I_ball * ω_ball
= 0 + I_ball * ω_ball
= I_ball * ω_ball
Setting the initial and final angular momenta equal to each other, we can solve for the angular velocity of the ball (ω_ball):
I_hoop * ω_hoop + I_ball * ω_ball = I_ball * ω_ball
Simplifying the equation:
I_hoop * ω_hoop = 0
Since ω_hoop is zero, we can solve for ω_ball:
I_ball * ω_ball = 0
So, ω_ball = 0.
This means that at the bottom of the hoop, the ball has zero angular velocity. Therefore, the ball is at rest at the bottom of the hoop.
Now, let's calculate the location along the x-axis (the position) at the bottom of the hoop.
The ball is initially at the rightmost side of the hoop, horizontally aligned with the hoop's center. We can define this point as x = 0 on the x-axis.
As the ball rolls down, it moves along the inside of the hoop, which has a radius of 1.4 m. At the bottom of the hoop, the ball has moved a distance equal to half the circumference of the hoop since it started at the top.
The circumference of the hoop is given by 2πR_hoop, so the distance the ball has traveled is πR_hoop.
Therefore, the location along the x-axis of the ball when it reaches the bottom of the hoop is πR_hoop, which is approximately 4.4 m.
So, the ball is located at approximately x = 4.4 m when it reaches the bottom of the hoop.
To calculate the location along the x-axis where the ball reaches the bottom of the hoop, we need to analyze the energy conservation and rotational dynamics of the system.
Let's consider the following variables:
- R_h: Radius of the hoop = 1.4 m
- m_h: Mass of the hoop = 11.5 kg
- R_b: Radius of the ball = R_h/4 = 1.4/4 = 0.35 m
- m_b: Mass of the ball = 2 * m_h = 2 * 11.5 kg = 23 kg
- x: Location along the x-axis where the ball reaches the bottom of the hoop (what we need to find)
First, let's find the moment of inertia for the hoop and the ball:
1. Moment of Inertia of the Hoop:
The moment of inertia of a hoop about its axis of rotation passing through its center is given by I_h = m_h * R_h^2.
2. Moment of Inertia of the Ball:
The moment of inertia of a solid sphere about its axis of rotation passing through its center is given by I_b = (2/5) * m_b * R_b^2.
The total moment of inertia for the system (hoop + ball) is calculated as:
I_total = I_h + I_b
Next, let's calculate the potential energy and the kinetic energy of the system at the initial and final positions:
1. Initial Position:
At the initial position, the ball is on the rightmost side of the hoop, horizontally aligned with the hoop's center. It has potential energy only.
Potential Energy at Initial Position:
PE_initial = m_b * g * R_h
2. Final Position:
At the final position, the ball is at the bottom of the hoop, having rolled down inside the hoop. It has both potential and kinetic energy.
Potential Energy at Final Position:
PE_final = 0 (the ball is at its lowest position)
Kinetic Energy at Final Position:
KE_final = (1/2) * I_total * ω^2
In this final position, since the ball is rolling without slipping, its translational velocity is related to its angular velocity by the equation: v = ω * R_b
Since the ball reaches the bottom of the hoop, its potential energy at the final position is equal to its initial potential energy. Therefore:
m_b * g * R_h = (1/2) * I_total * ω^2
Next, substitute the expressions for I_h and I_b, and ω = v / R_b, into the equation above to get:
m_b * g * R_h = (1/2) * (m_h * R_h^2 + (2/5) * m_b * R_b^2) * (v / R_b)^2
Rearrange the equation to solve for v:
v = sqrt( (2 * g * R_h) / (2/5 + m_h / m_b) )
Finally, substitute v = ω * R_b into the equation above to get:
ω = sqrt( (2 * g * R_h) / (2/5 + m_h / m_b) ) / R_b
The angular velocity ω gives us the rate of change of the angle θ of the hoop as it rolls down.
Since the ball rolls without slipping inside the hoop, the distance l traveled along the hoop is given by l = R_b * θ.
Now we can calculate the distance traveled by the ball inside the hoop:
l = R_b * θ = R_b * (2 * π) = 0.35 * (2 * π) = 0.35 * 2π = 2.2π
Lastly, since the x-axis is horizontal, we need to find the value of x for this distance traveled along the hoop:
x = l - R_h = 2.2π - 1.4
Thus, the location along the x-axis where the ball reaches the bottom of the hoop is approximately x = 0.778 m.