Solving Angular Speed of a Circular Disk

[teddygrams]

A flat uniform circular disk (radius = 4.69 m, mass = 280 kg) is initially stationary. The disk is free to rotate in the horizontal plane about a frictionless axis perpendicular to the center of the disk. A 55.0-kg person, standing 1.59 m from the axis, begins to run on the disk in a circular path and has a tangential speed of 1.70 m/s relative to the ground. Find the resulting angular speed (in rad/s) of the disk.

i know that i have to use :

0=I(person)w(final person) + I(disk)w(final disk)

and i know I=mr^2

but i just don't know how to rearrange everything...correctly..

 

[Nate810]

The angular speed of the disk is given by:w(final disk) = (I(person)w(final person)) / I(disk)where I(person) = 55.0 kg * 1.59 m^2 = 87.45 kgm^2I(disk) = 280 kg * 4.69 m^2 = 1319.2 kgm^2w(final person) = 1.70 m/sSubstituting in the above equation gives:w(final disk) = (87.45 kgm^2 * 1.70 m/s) / 1319.2 kgm^2 = 0.15 rad/s

 

[m2003]

I would first start by defining the variables in the problem. The moment of inertia (I) of the person and the disk can be calculated using the formula I=mr^2, where m is the mass and r is the radius. The final angular speed of the disk can be represented as w(final disk) and the final angular speed of the person can be represented as w(final person).

Next, I would use the conservation of angular momentum principle, which states that the total angular momentum of a system remains constant unless acted upon by an external torque. In this problem, the initial angular momentum of the system is zero since the disk is initially stationary. Therefore, we can set the initial angular momentum equal to the final angular momentum.

0=I(person)w(final person) + I(disk)w(final disk)

Now, we can substitute the values we know into the equation. The moment of inertia of the person can be calculated using the formula I=mr^2, where m is the mass of the person and r is the distance from the axis (1.59 m in this case). The moment of inertia of the disk is already given in the problem as 280 kg * 4.69 m^2. We can also substitute the value for the final tangential speed of the person (1.70 m/s) and solve for the final angular speed of the disk.

0=(55 kg * (1.59 m)^2)(w(final person)) + (280 kg * 4.69 m^2)(w(final disk))

Solving for w(final disk), we get:

w(final disk) = -(55 kg * (1.59 m)^2)(w(final person)) / (280 kg * 4.69 m^2)

Plugging in the values, we get:

w(final disk) = -0.032 rad/s

The negative sign indicates that the disk is rotating in the opposite direction of the person's motion. Therefore, the resulting angular speed of the disk is 0.032 rad/s.