As seen from above, a playground carousel is rotating counterclockwise about its center on frictionless bearings. A person standing still on the ground grabs onto one of the bars on the carousel very close to its outer edge and climbs aboard. Thus, this person begins with an angular speed of zero and ends up with a nonzero angular speed, which means that he underwent a counterclockwise angular acceleration. The carousel has a radius of 1.70 m, an initial angular speed of 3.42 rad/s, and a moment of inertia of 122 kg · m2. The mass of the person is 46.3 kg. Find the final angular speed of the carousel after the person climbs aboard.
To solve this problem, we can use the conservation of angular momentum. The initial angular momentum of the system is the sum of the angular momentum of the carousel and the angular momentum of the person. Since the person is initially not moving, their initial angular momentum is zero. The final angular momentum of the system is the sum of the angular momentum of the carousel and the person when both are moving together with the same angular velocity.
Let's denote the initial angular velocity of the carousel as ω1, the final angular velocity as ω2, the moment of inertia of the carousel as I, and the mass and radius of the carousel as m and R, respectively.
Initial angular momentum (L1) = Iω1 + (mass of person) * (angular momentum of person)
Since the person is initially not moving,
L1 = Iω1
Final angular momentum (L2) = Iω2 + (mass of person) * (angular momentum of person)
Since both the carousel and the person are moving with the same final angular velocity,
L2 = Iω2 + (mass of person) * (R^2 * ω2)
According to the conservation of angular momentum, the initial and final angular momenta should be equal:
L1 = L2
Iω1 = Iω2 + (mass of person) * (R^2 * ω2)
Now we can plug in the given values to find the final angular velocity ω2:
ω1 = 3.42 rad/s
I = 122 kg * m^2
mass of person = 46.3 kg
R = 1.70 m
Substituting these values into the equation, we get:
(122 kg * m^2) * (3.42 rad/s) = (122 kg * m^2) * ω2 + (46.3 kg) * ((1.70 m)^2) * ω2
Simplify and solve for ω2:
417.64 kg * m^2/s = (122 + 46.3 * 1.7^2) kg * m^2 * ω2
ω2 = 417.64 kg * m^2/s / (122 + 46.3 * 1.7^2) kg * m^2
ω2 ≈ 2.21 rad/s
The final angular speed of the carousel after the person climbs aboard is approximately 2.21 rad/s.
To find the final angular speed of the carousel, we can use the principle of conservation of angular momentum. Angular momentum is given by the equation:
L = Iω
Where L is the angular momentum, I is the moment of inertia, and ω is the angular speed.
Initially, the angular momentum of the carousel is given by:
L_initial = I_carousel * ω_initial
Final angular momentum is given by:
L_final = (I_carousel + I_person) * ω_final
Since the person grabs onto one of the bars close to the outer edge of the carousel, we can assume that the entire mass of the person is concentrated at the outer edge. Thus, the moment of inertia of the person can be calculated as:
I_person = m_person * r_person^2
Where m_person is the mass of the person and r_person is the radius from the center of the carousel to the outer edge where the person grabs onto.
Substituting the values given in the problem:
I_person = 46.3 kg * (1.70 m)^2 = 133.57 kg·m^2
Now, we can set up the conservation of angular momentum equation:
L_initial = L_final
I_carousel * ω_initial = (I_carousel + I_person) * ω_final
Substituting the given values:
122 kg·m^2 * 3.42 rad/s = (122 kg·m^2 + 133.57 kg·m^2) * ω_final
Now, solve for ω_final:
ω_final = (122 kg·m^2 * 3.42 rad/s) / (122 kg·m^2 + 133.57 kg·m^2)
ω_final ≈ 3.14 rad/s
Therefore, the final angular speed of the carousel after the person climbs aboard is approximately 3.14 rad/s.
To find the final angular speed of the carousel after the person climbs aboard, we can use the principle of conservation of angular momentum.
The initial angular momentum of the carousel can be calculated using the formula:
L1 = I1 * ω1
where L1 is the initial angular momentum, I1 is the moment of inertia of the carousel, and ω1 is the initial angular speed of the carousel.
Substituting the values given:
L1 = 122 kg · m^2 * 3.42 rad/s
L1 = 417.24 kg · m^2/s
When the person climbs aboard, their moment of inertia is added to the system. The moment of inertia of the person can be approximated by considering them as a point mass at the outer edge of the carousel:
I2 = m * r^2
where I2 is the moment of inertia of the person, m is the mass of the person, and r is the radius of the carousel.
Substituting the values given:
I2 = 46.3 kg * (1.7 m)^2
I2 = 139.59 kg · m^2
Now, with the person aboard, the final angular momentum of the system can be calculated as:
L_f = (I1 + I2) * ω_f
where L_f is the final angular momentum and ω_f is the final angular speed of the carousel.
Since angular momentum is conserved, we can set L1 equal to L_f:
L1 = (I1 + I2) * ω_f
Rearranging the equation to solve for ω_f:
ω_f = L1 / (I1 + I2)
Substituting the values:
ω_f = 417.24 kg · m^2/s / (122 kg · m^2 + 139.59 kg · m^2)
ω_f ≈ 1.362 rad/s
Therefore, the final angular speed of the carousel after the person climbs aboard is approximately 1.362 rad/s.