Question

A student sits on a rotating stool holding two 1 kg objects. When his arms are extended horizontally, the objects are 1.1 m from the axis of rotation, and he rotates with angular speed of 0.64 rad/sec. The moment of inertia of the student plus the stool is 4 kg m2 and is assumed to be constant. The student then pulls the objects horizontally to a radius 0.33 m from the rotation axis.

Required:
a. Find the new angular speed of the student.
b. Find the kinetic energy of the student before and after the objects are pulled in.

Answer 1

a) ω₁ = 0.97 rad/sec

b) K₀ = 1.31 J

Kf = 1.99 J

Step-by-step explanation:

a)

• Assuming no external torques acting on the student, total angular momentum must be conserved, as follows:

`L_(o) = L_(f) (1)`

• The angular momentum of a rigid body rotating respect an axis of rotation can be written as follows:

`L = I*\omega (2)`

• In order to get ωf, since ω₀ = 0.64 rad/sec, we need to find the values of the initial moment of inertia, I₀, and the final one, If:
• I₀ = 4 kg*m² + 1kg*(1.1 m)² + 1kg*(1.1m)² = 6.42 kg*m² (3)
• If = 4 kg*m² + 1kg*(0.33m)² + 1kg*(0.33m)² = 4.22 kg*m² (4)
• Replacing (3), (4) in (1) we can solve for ωf:

`\omega_(f) = (I_(o) *\omega_(o) )/(I_(f) ) = (6.42kgm2*0.64rad/sec)/(4.22kgm2) = 0.97 rad/sec (5)`

b)

• Since the student is not translating but he is only rotating, all his kinetic energy is rotational kinetic energy.
• The expression for the kinetic energy of a rotating rigid body, around an axis of rotation is as follows:

`K_(rot) = (1)/(2) * I * \omega^(2) (6)`

• The initial kinetic energy of the student, before the objects are pulled in, is as follows:

`K_(roto) = (1)/(2) *I_(o) * \omega_(o) ^(2) = (1)/(2)* 6.42kgm2*(0.64rad/sec)^(2) = 1.32 J (7)`

• The final kinetic energy is given by the following expression:
• 
  `K_(rotf) = (1)/(2) *I_(f) * \omega_(f) ^(2) = (1)/(2)* 4.22kg*m2*(0.97rad/sec)^(2) = 1.99 J (8)`