Ronald Johnson, who is 80.0 kg, is running at 4.00 m/s when he suddenly jumps onto a stationary playground merry-go-round at a distance 1.50 m from the axis of rotation of the merry-go-round. Ronald Johnson is traveling tangentially to the edge of the merry-go-round just before jumping on. The moment of inertia about its axis of rotation is 750 kg∙m2 and very little friction at its rotation axis. What is the angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it?
To solve this problem, we can use the conservation of angular momentum. The total angular momentum before Ronald jumps onto the merry-go-round must equal the total angular momentum after he jumps onto it.
Before Ronald jumps, the angular momentum of the merry-go-round (L1) is 0 (since it is stationary), and the angular momentum of Ronald (L2) is given by L2 = m * r * v, where m is his mass, r is the distance from the axis of rotation, and v is his velocity. So,
L2 = (80 kg) * (1.50 m) * (4.00 m/s) = 480 kg*m^2/s
The total initial angular momentum (L_initial) is the sum of the two: L_initial = L1 + L2 = 0 + 480 kg*m^2/s = 480 kg*m^2/s
After Ronald jumps onto the merry-go-round, the angular momentum of the system is given by the moment of inertia of the merry-go-round (I) multiplied by its angular velocity (ω). Therefore,
L_final = I * ω
Now using conservation of angular momentum, we set the initial and final angular momenta equal:
L_initial = L_final
480 kg*m^2/s = (750 kg*m^2 + 80 kg * (1.50 m)^2) * ω
Now, we solve for ω:
ω = 480 kg*m^2/s / (750 kg*m^2 + 80 kg * (1.50 m)^2)
ω = 480 kg*m^2/s / (750 kg*m^2 + 80 kg * 2.25 m^2)
ω = 480 kg*m^2/s / (750 kg*m^2 + 180 kg*m^2)
ω = 480 kg*m^2/s / 930 kg*m^2
ω ≈ 0.516 m^2/s
Therefore, the angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it is approximately 0.516 rad/s.
Well, let's start with Ronald Johnson, the jumping superstar! So, he's running at 4.00 m/s, ready to hop onto the merry-go-round. I hope he sticks the landing!
Now, when Ronald jumps, he lands a distance of 1.50 m from the axis of rotation. Let's not forget that he's traveling tangentially to the edge of the merry-go-round just before jumping on. So, we're not dealing with any fancy diagonal jumps here.
Now, the moment of inertia of the merry-go-round is given as 750 kg∙m². What a mouthful, huh? But hey, less friction means more fun on the merry-go-round!
To find the angular velocity of the merry-go-round just after Ronald jumps on, we'll use the principle of conservation of angular momentum. This principle states that the initial angular momentum of the system (Ronald + merry-go-round) will be equal to the final angular momentum.
Now, before Ronald jumps onto the merry-go-round, his angular momentum is given by:
Angular momentum = mass × velocity × radius
So, for Ronald Johnson:
Angular momentum = 80.0 kg × 4.00 m/s × 1.50 m
But once Ronald lands on the merry-go-round, his angular momentum will be transferred to the rotating merry-go-round. So we have:
Angular momentum of Ronald + Angular momentum of merry-go-round = Final angular momentum
We're looking for the angular velocity of the merry-go-round, so let's solve for that.
Final angular momentum = (Angular momentum of Ronald + Angular momentum of merry-go-round) / Moment of inertia of merry-go-round
Now, plug in the values we know and crunch those numbers to get the angular velocity. And hopefully, Ronald and the merry-go-round will have a fun-filled and spin-tacular time together!
To solve this problem, we can apply the principle of conservation of angular momentum.
1. Begin by calculating Ronald Johnson's initial angular momentum. The formula for angular momentum is given by L = I * ω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.
Given:
I (moment of inertia) = 750 kg∙m^2
ω (initial angular velocity) = ?
We can rearrange the formula to solve for ω: ω = L / I.
Since Ronald Johnson is initially traveling tangentially to the edge of the merry-go-round, his linear momentum is equal to the initial angular momentum. Linear momentum is given by p = m * v, where m is the mass and v is the velocity.
Given:
m (mass) = 80.0 kg
v (velocity) = 4.00 m/s
Linear momentum, p = m * v = 80.0 kg * 4.00 m/s = 320 kg∙m/s
Now we substitute the value of angular momentum (L), moment of inertia (I), and mass (m) into the formula to find the initial angular velocity (ω):
ω = L / I = p / (m * r), where r is the distance from the axis of rotation to the point of contact.
Given:
r (distance from the axis of rotation) = 1.50 m
ω = 320 kg∙m/s / (80.0 kg * 1.50 m) = 2.67 rad/s
Ronald Johnson's initial angular velocity just before jumping onto the merry-go-round is 2.67 rad/s.
2. Now, let's calculate the final angular velocity of the merry-go-round after Ronald Johnson jumps onto it.
The principle of conservation of angular momentum states that the total angular momentum before and after an event remains constant.
The initial angular momentum of Ronald Johnson is given by L_initial = I_initial * ω_initial.
The final angular momentum of Ronald Johnson and the merry-go-round is given by L_final = I_final * ω_final.
Since we know that the initial and final angular momenta are the same (conserved), we can write:
L_initial = L_final
Substituting the values we know, we get:
I_initial * ω_initial = I_final * ω_final
The moment of inertia of Ronald Johnson is negligible as compared to the moment of inertia of the merry-go-round. Therefore, we can approximate it as being zero.
Therefore, the equation becomes:
0 * ω_initial = I_final * ω_final
Solving for ω_final:
ω_final = 0 rad/s
The angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it is 0 rad/s.
To find the angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it, you can use the principle of conservation of angular momentum. The angular momentum of an object is the product of its moment of inertia and its angular velocity.
The initial angular momentum of Ronald can be calculated as follows:
L_initial = I_initial * ω_initial
Where:
L_initial : Initial angular momentum
I_initial : Initial moment of inertia of the merry-go-round
ω_initial : Initial angular velocity
Since Ronald is not initially on the merry-go-round, his initial angular momentum is zero (L_initial = 0) because he has no initial rotational motion.
When Ronald jumps onto the merry-go-round, the total angular momentum of the system remains conserved. Therefore, the final angular momentum of the system (which now includes Ronald and the merry-go-round) is equal to the initial angular momentum.
L_initial = L_final
0 = I_merry_go_round * ω_final + I_Ronald * ω_Ronald
Where:
I_merry_go_round : Moment of inertia of the merry-go-round
ω_final : Final angular velocity of the merry-go-round after Ronald jumps onto it
I_Ronald : Moment of inertia of Ronald (can be approximated to a point mass)
ω_Ronald : Initial angular velocity of Ronald (due to his tangential motion)
In this case, Ronald is considered a point mass since his distance from the rotation axis is given. The moment of inertia calculations are simplified for a point mass as I = m * r^2.
The equation can be rewritten as:
0 = I_merry_go_round * ω_final + (m_Ronald * r_Ronald^2) * ω_Ronald
To solve for the final angular velocity (ω_final), rearrange the equation:
ω_final = -(m_Ronald * r_Ronald^2) * ω_Ronald / I_merry_go_round
Now, plug in the given values:
m_Ronald = 80.0 kg (mass of Ronald)
r_Ronald = 1.50 m (distance of Ronald from the rotation axis)
ω_Ronald = 4.00 m/s (initial angular velocity of Ronald due to his tangential motion)
I_merry_go_round = 750 kg∙m2 (moment of inertia of the merry-go-round)
Plug the values into the equation:
ω_final = -(80.0 kg * (1.50 m)^2 * 4.00 m/s) / 750 kg∙m^2
Now, calculate the value of ω_final:
ω_final = -0.48 rad/s
The negative sign indicates that the merry-go-round rotates in the opposite direction to the initial motion of Ronald.
Therefore, the angular velocity of the merry-go-round just after Ronald Johnson has jumped onto it is approximately -0.48 rad/s.