a 32 kg child stands at the edge of a circular platform that rotates at w=2rad/s about its center. Mass of platform=60kg, radius=2m.
The child walks from the rim towards the center. Find the Angular speed when the child reaches a point 0.500 m from the center.
The law of conservation of angular momentum
L₁ = L₂
{(MR²/2) +(mR²)}ω₁ ={(MR²/2) +(mr²)}ω₂
ω₂=ω₁{(MR²/2) +(mR²)}/{(MR²/2) +(mr²)}=
=2{60•4/2 + 32•4}/(60•4/2+ 32•0.25}=
=3.875 rad/s
Thank you!
To find the angular speed when the child reaches a point 0.500 m from the center, we can use the principle of conservation of angular momentum.
The angular momentum of the child on the rotating platform is given by:
L1 = I1 * ω1,
where L1 is the angular momentum, I1 is the moment of inertia, and ω1 is the initial angular speed.
The moment of inertia of the platform can be calculated using the formula:
I_platform = 0.5 * m_platform * r_platform^2,
where m_platform is the mass of the platform and r_platform is its radius.
Substituting the given values:
I_platform = 0.5 * 60 kg * (2 m)^2 = 240 kg*m^2.
The moment of inertia of the child can be calculated using the formula:
I_child = m_child * r_child^2,
where m_child is the mass of the child and r_child is the distance of the child from the center.
Substituting the given values:
I_child = 32 kg * (0.500 m)^2 = 8 kg*m^2.
The initial angular momentum of the child on the platform is:
L1 = I1 * ω1 = (I_platform + I_child) * ω1.
When the child reaches a point 0.500 m from the center, the moment of inertia of the child changes to:
I2 = m_child * r_2^2,
where r_2 is the new distance of the child from the center.
Substituting the given value:
I2 = 32 kg * (0.500 m)^2 = 8 kg*m^2.
The final angular momentum of the child on the platform is:
L2 = I2 * ω2,
where ω2 is the final angular speed.
According to the conservation of angular momentum, L1 = L2:
(I_platform + I_child) * ω1 = I2 * ω2.
Simplifying the equation:
(240 kg*m^2 + 8 kg*m^2) * ω1 = 8 kg*m^2 * ω2,
248 kg*m^2 * ω1 = 8 kg*m^2 * ω2.
Dividing both sides by 8 kg*m^2:
31 kg*m^2 * ω1 = ω2.
Since ω1 = 2 rad/s (given), we can substitute this value:
31 kg*m^2 * 2 rad/s = ω2.
Simplifying the equation:
ω2 = 62 rad/s.
Therefore, the angular speed when the child reaches a point 0.500 m from the center is 62 rad/s.
To solve this problem, we can use the principle of conservation of angular momentum.
Angular momentum (L) is given by the formula L = Iω, where I is the moment of inertia and ω is the angular velocity.
The moment of inertia of the circular platform is given by I = mr², where m is the mass of the platform and r is the radius of the platform.
First, let's calculate the moment of inertia of the platform:
I = m * r²
= 60 kg * (2 m)²
= 240 kg * m²
The initial angular momentum of the system (child + platform) is equal to the final angular momentum of the system, since there is no external torque acting on the system.
L_initial = L_final
(I_child + I_platform) * ω_initial = (I_child + I_platform) * ω_final
Since the child is initially at the rim of the platform, 0.5 m from the center, the moment of inertia of the child can be approximated as a point mass at the rim, given by I_child = m_child * r_child², where m_child is the mass of the child and r_child is the distance of the child from the center.
Let's calculate the moment of inertia of the child at the rim:
I_child = m_child * r_child²
= 32 kg * (2 m)²
= 128 kg * m²
The initial angular momentum of the system is therefore:
L_initial = (I_child + I_platform) * ω_initial
When the child moves towards the center, the moment of inertia of the child changes. At a point 0.5 m from the center, the moment of inertia can be calculated as:
I_child_final = m_child * r_final²
= 32 kg * (0.5 m)²
= 8 kg * m²
At this point, the final angular momentum of the system is:
L_final = (I_child_final + I_platform) * ω_final
Since angular momentum is conserved, we set the initial and final angular momenta equal to each other:
(I_child + I_platform) * ω_initial = (I_child_final + I_platform) * ω_final
Now, we can solve for ω_final, which is the angular speed of the platform when the child reaches a point 0.5 m from the center.
(128 kg * m² + 240 kg * m²) * 2 rad/s = (8 kg * m² + 240 kg * m²) * ω_final
Simplifying the equation:
368 kg * m² * rad/s = 248 kg * m² * ω_final
Dividing both sides by (248 kg * m²):
ω_final = (368 kg * m² * rad/s) / (248 kg * m²)
Now, we can calculate ω_final:
ω_final ≈ 1.48 rad/s
Therefore, the angular speed of the platform when the child reaches a point 0.5 m from the center is approximately 1.48 rad/s.