Question

A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 1 m from the axis of rotation, and he rotates with angular speed of 0.76 rad/sec.Calculate the final angular speed of the student.

Answer 1

The angular speed of the student is 1.144 rad/s.

Step-by-step explanation:

Given that,

Mass of two objects = 1 kg

Distance = 1 m

Angular speed = 0.76 rad/s

Suppose The moment of inertia of the student plus stool is 3.0 kg-m² and is assumed to be constant. The student then pulls the objects horizontally to 0.40 m from the rotation axis.

We need to calculate the initial moment of inertia of the student & stool plus the two masses:

Using formula of moment of inertia

`I_(i)=I_(s+s)+mr^2+mr^2`

Put the value into the formula

`I_(i)=3.0+1.0*1.0^2+1.0*1.0^2`

`I_(i)=5\ kg-m^2`

We need to calculate the the final moment of inertia of the student & stool plus the two masses

Using formula of moment of inertia

`I_(f)=I_(s+s)+mr^2+mr^2`

Put the value into the formula

`I_(f)=3.0+1.0*0.40^2+1.0*0.40^2`

`I_(f)=3.32\ kg-m^2`

We need to calculate the angular speed

Using conservation of momentum

`I_(i)*\omega_(i)=I_(f)*\omega_(f)`

Put the value into the formula

`5*0.76=3.32*\omega_(f)`

`\omega_(f)=(5*0.76)/(3.32)`

`\omega_(f)=1.144\ rad/s`

Hence,The angular speed of the student is 1.144 rad/s.