A student sits on a rotating stool holding two 2 kg objects. When his arms are extended horizontally, the objects are 0.9 m from the axis of rotation, and he rotates with angular speed of 0.74 rad/sec. The moment of inertia of the student plus the stool is 7 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.28 m from the rotation axis.

(a) Find the new angular speed of the student.
rad/s

(b) Find the kinetic energy of the student before and after the objects are pulled in.

before: J
after: J

A) ω2 = 1.394 rad/s

B) KE_i = 1.7085 J and KE_f = 3.2196J

Explanation:

We are given;

Mass of objects: m1 = m2 = 2kg

Radius: r1 = 0.9m ; r2 = 0.28m

Angular velocity: ω1 = 0.74 rad/s

Moment of inertia of stool and object: I = 3kg.m²

A) Using the law of conservation of angular momentum:

I1•ω1 = I2•ω2

The values of the moment of inertia according to the parallel axis theorem will be:

I1 = I + 2mr1²

So, I1 = 3 + 2(2 x 0.9²)

I1 = 6.24 kg.m²

Similarly,

I2 = I + 2mr2²

I2 = 3 + 2(2)(0.28²)

I2 = 3.3136 kg.m²

Since I1•ω1 = I2•ω2

Thus, ω2 = I1•ω1/I2

ω2 = 6.24 x 0.74/3.3136

ω2 = 1.394 rad/s

B) The initial kinetic energy is:

KE_i = (1/2)•I1•ω1²

KE_i = (1/2) x 6.24 x 0.74²

KE_i = 1.7085 J

The final kinetic energy is;

KE_f = (1/2)•I2•ω2²

KE_f = (1/2) x 3.3136 x 1.394²

KE_f = 3.2196J

it has to be b, about 97 percent.

a=d/t

a="triangle"v/t

explanation:

a = acceleration