A horizontal plat form in the shap 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 student has reached at point 0.5m from the center
Let's first find the moment of inertia (I) of the system. The moment of inertia of a disk rotating about its center is given by 1/2mr^2. In this case, the student is not at the center, so we need to use the parallel axis theorem to find the moment of inertia with respect to the center. The parallel axis theorem states that I = Icm + md^2, where Icm is the moment of inertia about the center of mass and d is the distance between the center of mass and the axis of rotation.
At the beginning when the student is at the edge, the moment of inertia is:
I = 1/2mr^2 + m(2 - r)^2 = 1/2(100)(2)^2 + (60)(2-2)^2 = 280 kg m^2
When the student walks to a distance of 0.5m from the center, the distance d between the center of mass and the axis of rotation becomes 2 - 0.5 = 1.5m. Therefore, the moment of inertia becomes:
I' = 1/2mr^2 + m(2 - 1.5)^2 = 1/2(100)(2)^2 + (60)(0.5)^2 = 250 kg m^2
The conservation of angular momentum and the law of conservation of energy can be used to solve this problem.
Let's apply the law of conservation of energy first. The total energy of the system at the beginning is:
E = 1/2Iω^2 = 1/2(280)(2)^2 = 560 J
where ω is the angular velocity at the beginning.
As the student walks towards the center, the total energy of the system is conserved. The kinetic energy of the system is given by:
K = 1/2I'ω'^2 + 1/2m(vcm')^2
where ω' is the angular velocity when the student is at a distance of 0.5m from the center, and vcm' is the speed of the center of mass of the system at this point. Since the system is rotating about a fixed axis, the center of mass moves with a constant velocity. Therefore, vcm' is equal to the tangential velocity of the student with respect to the center, which is given by vcm' = ω'r.
The potential energy of the system decreases as the student walks towards the center. At the beginning, the potential energy of the system is:
U = mgh = 60(9.81)(2) = 1177.2 J
where h is the height of the student above the ground. When the student is at a distance of 0.5m from the center, the potential energy of the system is:
U' = mgh' = 60(9.81)(1.5) = 882.9 J
The total energy at this point is:
E' = K + U' = 1/2I'ω'^2 + 1/2m(ω'r)^2 + mgh' = 1/2(250)ω'^2 + 1/2(60)(ω')^2 + 882.9
Since the total energy is conserved, we have E = E', which gives:
1/2(280)(2)^2 = 1/2(250)ω'^2 + 1/2(60)(ω')^2 + 882.9
Simplifying this equation, we get:
125ω'^2 + 30(ω')^2 = 713.6
155ω'^2 = 713.6
ω' = sqrt(4.6) = 2.14 rad/s
Therefore, the angular velocity when the student is at a distance of 0.5m from the center is 2.14 rad/s.
The initial r of the student is 2, not ZERO
the height does not change
What happened to angular momentum ?
I apologize for the confusion. If the initial distance of the student from the center is 2m and the height remains the same throughout, then the application of the law of conservation of energy remains the same as shown above. However, the angular momentum is not conserved in this case because an external torque acts on the system due to the movement of the student towards the center.
The initial angular momentum of the system is:
L = Iω = (1/2mr^2 + md^2)ω = (1/2)(100)(2)^2(2) = 400 kg m^2/s
When the student moves towards the center, the moment of inertia of the system changes, but the external torque due to the movement of the student remains constant and equal to the force exerted by the student towards the center multiplied by the distance of the student from the center (which is 2m at the beginning). This torque is given by:
τ = Fd = mgd = (60)(9.81)(2) = 1177.2 Nm
The change in angular momentum of the system due to the external torque is:
ΔL = τΔt
where Δt is the time it takes for the student to move from the edge to a distance of 0.5m from the center. Since the student moves slowly, we can assume that the change in angular momentum is small and happens over a relatively long period of time. Therefore, we can approximate Δt as the time it takes for the student to move a distance of 1.5m (from 2m to 0.5m) with a constant speed of v = rΔθ/Δt, where Δθ is the change in angle covered by the student during this time. This gives:
Δt = Δθ/ω = (1.5/2)/2 = 0.375 s
Therefore, the change in angular momentum of the system is:
ΔL = τΔt = (1177.2)(0.375) = 441.9 kg m^2/s
The final angular momentum of the system is:
L' = I'ω'
where I' is the moment of inertia of the system when the student is at a distance of 0.5m from the center and ω' is the angular velocity at this point. We have already calculated I' and ω' above as 250 kg m^2 and 2.14 rad/s, respectively. Substituting these values, we get:
L' = I'ω' = (250)(2.14) = 535 kg m^2/s
Therefore, the change in angular momentum of the system due to the external torque is equal to the difference between the final and initial angular momentum of the system:
ΔL = L' - L = 535 - 400 = 135 kg m^2/s
This confirms that the angular momentum is not conserved in this case due to the external torque caused by the movement of the student towards the center.
Nonsense
There is no external torque on the student - disc system
There is no change in the angular momentum of the student - disc system
Try the spinning skater pulling her arms in !
You are correct. I apologize for my mistake in the previous response.
In this case, there is no external torque acting on the system, and therefore, the angular momentum remains conserved. As the student moves towards the center, the moment of inertia of the system changes, but the angular velocity of the system also changes to keep the angular momentum constant.
The initial angular momentum of the system is:
L = Iω = (1/2mr^2 + md^2)ω = (1/2)(100)(2)^2(2) = 400 kg m^2/s
When the student moves towards the center, the moment of inertia of the system changes, and the angular velocity of the system changes such that the angular momentum remains constant. The final moment of inertia of the system is:
I' = 1/2mr'^2 + md'^2 = (1/2)(100)(0.5)^2 + (60)(1.5)^2 = 290 kg m^2
where r' is the distance of the student from the center when he reaches at a distance of 0.5m from the center and d' is the distance between the center of mass and the axis of rotation. The final angular velocity of the system can be calculated using the conservation of angular momentum:
L = L' = I'ω'
where ω' is the final angular velocity of the system. Solving for ω', we get:
ω' = L/I' = 400/290 = 1.38 rad/s
Therefore, the final angular velocity of the system when the student reaches at a distance of 0.5m from the center is 1.38 rad/s. Again, I apologize for my previous mistake and any confusion it may have caused.
Whew !
I did not check the arithmetic but at least the approach is correct.