A horizontal plat form in the shape of a circular disk rotates in a horizontal plane about a frictionless vertical axle. The plat form has Mass m=100kg and radius r=2m. A students whose mass m=60kg walks slowly on the plat form towards the center. If the angular velocity of the system is 2rad/s when the students at the edge. Then calculate the angular velocity when the students has reached at point 0.5m from the centre
The conservation of angular momentum can be used to solve this problem:
L_i = L_f
where
L_i = initial angular momentum of the system
L_f = final angular momentum of the system
Initially, the platform is rotating with an angular velocity of 2 rad/s and no external torques are acting on the system. Therefore,
L_i = I * w_i
where
I = moment of inertia of the platform
w_i = initial angular velocity of the platform
The moment of inertia of a circular disk is I = (1/2) * m * r^2. Substituting the given values,
L_i = (1/2) * 100 kg * (2 m)^2 * 2 rad/s
= 400 kg m^2/s
When the student reaches a point 0.5 m from the center, the moment of inertia of the system changes because of the redistribution of mass. However, the angular momentum of the system remains conserved. Therefore,
L_i = L_f
I_i * w_i = I_f * w_f
where
I_i = initial moment of inertia of the system (with the student at the edge)
w_i = initial angular velocity of the system
I_f = final moment of inertia of the system (with the student at 0.5 m from the center)
w_f = final angular velocity of the system (to be determined)
The moment of inertia of the system with the student at the edge is I_i = (1/2) * (100+60) kg * (2 m)^2 = 320 kg m^2. The moment of inertia of the system with the student at 0.5 m from the center can be calculated using the parallel axis theorem as follows:
I_f = I_c + m * d^2
where
I_c = moment of inertia of the disk about its own center of mass (i.e., without the student)
m = mass of the student
d = distance between the center of the disk and the student's location
The moment of inertia of the disk about its own center of mass is I_c = (1/2) * 100 kg * (2 m)^2 = 200 kg m^2. Substituting the given values,
I_f = 200 kg m^2 + 60 kg * (0.5 m)^2
= 215 kg m^2
Substituting all the known values into the conservation of angular momentum equation and solving for w_f,
I_i * w_i = I_f * w_f
320 kg m^2 * 2 rad/s = 215 kg m^2 * w_f
w_f = 2.37 rad/s (rounded to two significant figures)
Therefore, the angular velocity of the system when the student has reached a point 0.5 m from the center is 2.37 rad/s.
To calculate the angular velocity when the student is at a point 0.5m from the center, we can use the principle of conservation of angular momentum.
The angular momentum of the system is conserved. Initially, when the student is at the edge, the angular momentum of the system is:
L₁ = (mass of platform) × (radius of platform) × (angular velocity)
L₁ = (100kg) × (2m) × (2rad/s)
L₁ = 400kg·m²/s
When the student moves towards the center, the moment of inertia of the system changes. The moment of inertia of the platform can be calculated as:
I = (mass of platform) × (radius of platform)²
I = (100kg) × (2m)²
I = 400kg·m²
The moment of inertia of the system consisting of the platform and the student can be calculated as:
I' = (moment of inertia of platform) + (moment of inertia of student)
I' = I + (mass of student) × (distance from center)²
I' = 400kg·m² + (60kg) × (0.5m)²
I' = 400kg·m² + 15kg·m²
I' = 415kg·m²
Now, we can use the principle of conservation of angular momentum again:
L₁ = L₂
(mass of platform) × (radius of platform) × (angular velocity₁) = (mass of platform + mass of student) × (distance from center) × (angular velocity₂)
(100kg) × (2m) × (2rad/s) = (100kg + 60kg) × (0.5m) × (angular velocity₂)
400kg·m²/s = 160kg × m × rad/s
Now, we can solve for angular velocity₂:
angular velocity₂ = (400kg·m²/s) / (160kg × m × rad/s)
angular velocity₂ = 2.5 rad/s
Therefore, the angular velocity when the student has reached a point 0.5m from the center is 2.5 rad/s.