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

Answer:

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

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

$I_{i}=3.0+1.0\times1.0^2+1.0\times1.0^2$

$I_{i}=5\ kg-m^2$

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

$I_{f}=3.0+1.0\times0.40^2+1.0\times0.40^2$

$I_{f}=3.32\ kg-m^2$

We need to calculate the angular speed

Using conservation of momentum

$I_{i}\times\omega_{i}=I_{f}\times\omega_{f}$

Put the value into the formula

$5\times0.76=3.32\times\omega_{f}$

$\omega_{f}=\dfrac{5\times0.76}{3.32}$

$\omega_{f}=1.144\ rad/s$

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