1.what is the final velocity of a hoop that rolls with out slipping down a 3.00m high hill starting at rest?
2. a 1.80 m radius playground merry go round (disk = 1/2 m r^2) hass a mass of 120 kg and is rotating with an angular velocity of .600 rev/s. what is the angular velocity after a 28 kg kid gets on to it by grabbing the outer edge? the child is initially at rest
1. V^2 = Vo^2 + 2g*h
V^2 = 0 + 19.6*3 = 58.8
V = 7.67 m/s.
2.
To solve these problems, we can use the concepts of conservation of energy and conservation of angular momentum.
1. The final velocity of a hoop that rolls without slipping down a hill can be found by considering the conservation of energy. The initial potential energy is converted into kinetic energy at the bottom of the hill.
The potential energy at the top of the hill (initial) is given by mgh, where m is the mass of the hoop, g is the acceleration due to gravity, and h is the height of the hill.
The kinetic energy at the bottom of the hill (final) is given by (1/2)mv^2, where v is the final velocity.
Since the hoop is rolling without slipping, we can relate the linear velocity v and the angular velocity ω using the equation v = ωr, where r is the radius of the hoop.
Equating the initial potential energy and the final kinetic energy, we have:
mgh = (1/2)mv^2
Simplifying and canceling out the mass, we get:
gh = (1/2)v^2
Solving for v, we have:
v = sqrt(2gh)
Substituting the given values:
g = 9.8 m/s^2
h = 3.00 m
v = sqrt(2 * 9.8 m/s^2 * 3.00 m)
= sqrt(58.8 m^2/s^2)
= 7.67 m/s
Therefore, the final velocity of the hoop is 7.67 m/s.
2. The angular momentum of the system remains constant before and after the child gets on the merry-go-round.
The formula for angular momentum is given by L = Iω, where L represents angular momentum, I represents moment of inertia, and ω represents angular velocity.
The moment of inertia for a disk is given by I = (1/2)m*r^2, where m is the mass of the disk and r is its radius.
Before the child gets on the merry-go-round, the total angular momentum is given by:
L_initial = I_initial * ω_initial
Substituting the given values:
I_initial = (1/2) * 120 kg * (1.80 m)^2
ω_initial = 0.600 rev/s = 0.600 * 2π rad/s (since 1 revolution = 2π radians)
Calculating I_initial and ω_initial:
I_initial = (1/2) * 120 kg * (1.80^2 m^2)
= 194.4 kg m^2
ω_initial = 0.600 * 2π rad/s
= 3.77 rad/s
L_initial = 194.4 kg m^2 * 3.77 rad/s
= 733.43 kg m^2/s
After the child gets on the merry-go-round by grabbing the outer edge, the moment of inertia changes. Now, the total angular momentum is given by:
L_final = I_final * ω_final
The disk's moment of inertia does not change as it is rotating, but the addition of the child changes the total moment of inertia.
The total moment of inertia is given by:
I_final = I_disk + I_child
Substituting the given values:
I_disk = (1/2) * 120 kg * (1.80 m)^2
I_child = m_child * r^2, where m_child is the mass of the child and r is the radius of the disk.
Calculating I_disk, I_child, and I_final:
I_disk = (1/2) * 120 kg * (1.80^2 m^2)
= 194.4 kg m^2
I_child = 28 kg * (1.80 m)^2
= 907.2 kg m^2
I_final = 194.4 kg m^2 + 907.2 kg m^2
= 1101.6 kg m^2
We can now solve for the final angular velocity by rearranging the equation:
L_final = I_final * ω_final
ω_final = L_final / I_final
Substituting the calculated values:
ω_final = 733.43 kg m^2/s / 1101.6 kg m^2
≈ 0.666 rad/s
Therefore, the angular velocity after the child gets on the merry-go-round is approximately 0.666 rad/s.
To find the final velocity of the hoop in question 1, we can use the principle of conservation of energy. The initial potential energy of the hoop at the top of the hill is equal to its final kinetic energy.
1. First, calculate the potential energy of the hoop at the top of the hill:
Potential energy = mass * acceleration due to gravity * height
We know the mass of the hoop and the height of the hill, but we need to determine the acceleration due to gravity. The standard value is approximately 9.8 m/s^2.
2. Next, find the final kinetic energy of the hoop:
Since the hoop is rolling without slipping, its final kinetic energy is a combination of rotational and translational energy:
Kinetic energy = (1/2) * moment of inertia * (angular velocity)^2 + (1/2) * mass * (velocity)^2
The moment of inertia for a solid hoop can be calculated using I = 1/2 * m * r^2, where m is the mass of the hoop, and r is the radius.
3. Equating the initial potential energy to the final kinetic energy will allow us to solve for the final velocity of the hoop.
For question 2, we need to use the principle of conservation of angular momentum. Since there is no external torque acting on the system, the angular momentum is conserved:
Initial angular momentum = Final angular momentum
The initial angular momentum is just the product of the initial moment of inertia and angular velocity. The final angular momentum is the sum of the initial angular momentum and the angular momentum added by the child.
1. Calculate the initial angular momentum using the initial angular velocity and moment of inertia of the merry-go-round.
2. Calculate the angular momentum added by the child using the mass of the child, the radius of the merry-go-round, and the angular velocity of the child after grabbing the outer edge.
3. Finally, find the final angular velocity by dividing the final angular momentum by the moment of inertia of the merry-go-round with the child on it.
Remember to convert angles from revolutions to radians when necessary, as 1 revolution is equal to 2π radians.