Question

A man stands on a platform that is rotating (without friction) with an angular speed of 2.00 rev/s; his arms are outstretched and he holds a brick in each hand. The rotational inertia of the system consisting of the man, bricks, and platform about the central axis is 7.55 kg·m2. If by moving the bricks the man decreases the rotational inertia of the system to 1.22 kg·m2, (a) what is the resulting angular speed of the platform and (b) what is the ratio of the new kinetic energy of the system to the original kinetic energy?

Answer 1

77.76729 rad/s or 12.37704 rev/s

`(K_n)/(K_o)=6.18852`

Step-by-step explanation:

`I_1` = Rotational inertia of the system consisting of the man, bricks, and platform about the central axis = 7.55 kgm²

`I_2` = Decreased the rotational inertia of the system = 1.22 kgm²

`\omega_1` = Angular velocity of the old system =
`4\pi`

`\omega_1` = Angular velocity of the new system

`K_n` = Kinetic energy of the new system

`K_o` = Kinetic energy of the old system

Here, angular momentum is conserved

`L_1=L_2\\\Rightarrow I_1\omega_1=I_2\omega_2\\\Rightarrow \omega_2=(I_1\omega_1)/(I_2)\\\Rightarrow \omega_2=(7.55* 2* 2\pi)/(1.22)\\\Rightarrow \omega_2=77.76729\ rad/s\\ =((7.55* 2* 2\pi)/(1.22))/(2\pi)=12.37704\ rev/s`

The resulting angular speed of the platform is 77.76729 rad/s or 12.37704 rev/s

`(K_n)/(K_o)=((1)/(2)I_2\omega_2^2)/((1)/(2)I_1\omega_1^2)\\\Rightarrow (K_n)/(K_o)=(I_2\omega_2^2)/(I_1\omega_1^2)\\\Rightarrow (K_n)/(K_o)=(1.22* 77.76729^2)/(7.55* (2* 2\pi)^2)\\\Rightarrow (K_n)/(K_o)=6.18852`

The ratio of the new kinetic energy of the system to the original kinetic energy is
`(K_n)/(K_o)=6.18852`