Question

A person with mass mp = 75 kg stands on a spinning platform disk with a radius of R = 1.59 m and mass md = 186 kg. The disk is initially spinning at ω = 2 rad/s. The person then walks 2/3 of the way toward the center of the disk (ending 0.53 m from the center). 1.What is the total moment of inertia of the system about the center of the disk when the person stands on the rim of the disk? 2.What is the total moment of inertia of the system about the center of the disk when the person stands at the final location 2/3 of the way toward the center of the disk? 3.What is the final angular velocity of the disk? 4.What is the change in the total kinetic energy of the person and disk? (A positive value means the energy increased.) 5.What is the centripetal acceleration of the person when she is at R/3? 6.If the person now walks back to the rim of the disk, what is the final angular speed of the disk?

Answer 1

Answer:

1. $I_0=424.7208\,kg.m^2$

2. $I_f=256.1808\,kg.m^2$

3. $\omega_f=3.3158\,rad.s^{-1}$

4. $\Delta KE=558.8432\,J$

5. $a_c=5.8271\,m.s^{-2}$

6.  $\omega_0=2\,rad.s^{-1}$

Explanation:

Given:

mass of the person, $m_p=75\,kg$

radius of the disk, $r=1.59\,m$

mass of the disk, $m_d=186\,kg$

initial angular speed of the disk, $\omega_0=2\,rad.s^{-1}$

distance of the person from the center of the disk, $d=0.53\,m$

1.

Initial moment of inertia of the system when the man stands at the rim of disk:

Moment of inertia of the disc:

$I_D=\frac{1}{2} .m_d.R^2$

$I_D=\frac{1}{2} \times 186\times 1.59^2$

$I_D=235.1133\,kg.m^2$

Now for the person, we treat the mass to be a point revolving around R:

$I_P=m_p.R^2$

$I_P=75\times 1.59^2$

$I_P=189.6075\,kg.m^2$

∴We have the moment of inertia of the system in this case as:

$I_0=I_D+I_P$

$I_0=235.1133+189.6075$

$I_0=424.7208\,kg.m^2$

2.

Moment of inertia when the person stands at 0.53 m from the center of the disk:

Moment of inertia of the disk will be constant:

$I_D=235.1133\,kg.m^2$

For the person, we treat the mass to be a point revolving around radius 0.53 m:

$I_P=75\times 0.53^2$

$I_P=21.0675\,kg.m^2$

∴We have the moment of inertia of the system

$I_f=I_D+I_P$

$I_f=235.1133+21.0675$

$I_f=256.1808\,kg.m^2$

3.

The final angular velocity of the disk:

Using the conservation of angular momentum:

$I_0.\omega_0=I_f.\omega_f$

$424.7208\times 2=256.1808\times \omega_f$

$\omega_f=3.3158\,rad.s^{-1}$

4.

Change in Kinetic Energy:

∵$KE=\frac{1}{2} I.\omega^2$

∴$\Delta KE= \frac{1}{2} (I_f.\omega_f^2-I_0.\omega_0^2)$

$\Delta KE=\frac{1}{2} (256.1808\times 3.3158^2-424.7208\times 2^2)$

$\Delta KE=558.8432\,J$

5.

Centripetal acceleration of the person when she is at R/3:

Centripetal acceleration is given as:

$a_c=r'.\omega^2$

we have ω=3.3158 radian per second at R=0.53 m

$a_c=\frac{R}{3} .\omega^2$

$a_c=0.53\times 3.3158^2$

$a_c=5.8271\,m.s^{-2}$

6.

If the person now walks back to the rim of the disk:

Then by the law of conservation of angular momentum the initial angular speed of  $\omega_0=2\,rad.s^{-1}$ will be restored.