A large horizontal circular platform (M=143.1 kg, r=3.01 m) rotates about a frictionless vertical axle. A student (m=75.3 kg) walks slowly from the rim of the platform toward the center. The angular velocity ω of the system is 2.10 rad/s when the student is at the rim. Find ω (in rad/s) when the student is 2.89 m from the center.
Physics - Damon, Tuesday, February 11, 2014 at 6:25pm
conservation of angular momentum (the figure skater spins faster when she pulls her arms in
I1 omega1 = I2 omega2
I of platform = (1/2)(143.1)(3.01)^2
original I of student = m r^2 = 75.3(3.01)^2
final I of student = 75.3(2.89)^2
Physics - Molly, Tuesday, February 11, 2014 at 6:54pm
something is wrong in the answer, can you explain it again please?..
Physics - Molly, Tuesday, February 11, 2014 at 6:54pm
something is wrong in the answer, can you explain it again please?..
Physics - Damon, Tuesday, February 11, 2014 at 7:15pm
Ok,
Original I = (1/2)(143.1)(3.01^2 + 75.3*(3.01)^26
= 648 + 682 = 1330 kg m^2
Final I = 648 + 75.3(2.89)^2 = 1276
1330 (2.1) = 1276 omega
so
omega = 2.19 radians/second
Sure, let's go through the problem again step by step to find the correct answer.
We have a large circular platform with a mass of 143.1 kg and a radius of 3.01 m. The student has a mass of 75.3 kg. The system is initially rotating with an angular velocity of 2.10 rad/s when the student is at the rim. We want to find the angular velocity when the student is 2.89 m from the center.
To solve this problem, we can use the conservation of angular momentum. The total angular momentum of the system before and after the student moves can be equated.
The initial angular momentum of the system (with the student at the rim) can be calculated as follows:
L1 = I_platform * ω1 + I_student * ω1
Where I_platform is the moment of inertia of the platform, I_student is the moment of inertia of the student, and ω1 is the initial angular velocity.
The moment of inertia of a circular platform rotating about a central axis is given by:
I_platform = (1/2) * M * r^2
Substituting the given values, we get:
I_platform = (1/2) * 143.1 * (3.01)^2
Now we need to calculate the moment of inertia of the student. The moment of inertia of a point mass rotating about an axis at a distance r from the center is given by:
I_point mass = m * r^2
Substituting the given values, we get the initial moment of inertia of the student:
I_student = 75.3 * (3.01)^2
The final angular momentum of the system (with the student at 2.89 m from the center) can be calculated as follows:
L2 = I_platform * ω2 + I_student * ω2
We want to find ω2, so we can rearrange the equation:
L2 = (I_platform + I_student) * ω2
Now, we can equate L1 and L2 to find ω2:
I_platform * ω1 + I_student * ω1 = (I_platform + I_student) * ω2
Plugging in the values of I_platform, I_student, and ω1, we can solve for ω2.