Answer:
a) ω₁ = 0.97 rad/sec
b) K₀ = 1.31 J
Kf = 1.99 J
Explanation:
a)
- Assuming no external torques acting on the student, total angular momentum must be conserved, as follows:
[tex]L_{o} = L_{f} (1)[/tex]
- The angular momentum of a rigid body rotating respect an axis of rotation can be written as follows:
[tex]L = I*\omega (2)[/tex]
- 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:
[tex]\omega_{f} = \frac{I_{o} *\omega_{o} }{I_{f} } = \frac{6.42kgm2*0.64rad/sec}{4.22kgm2} = 0.97 rad/sec (5)[/tex]
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:
[tex]K_{rot} = \frac{1}{2} * I * \omega^{2} (6)[/tex]
- The initial kinetic energy of the student, before the objects are pulled in, is as follows:
[tex]K_{roto} = \frac{1}{2} *I_{o} * \omega_{o} ^{2} = \frac{1}{2}* 6.42kgm2*(0.64rad/sec)^{2} = 1.32 J (7)[/tex]
- The final kinetic energy is given by the following expression:
- [tex]K_{rotf} = \frac{1}{2} *I_{f} * \omega_{f} ^{2} = \frac{1}{2}* 4.22kg*m2*(0.97rad/sec)^{2} = 1.99 J (8)[/tex]
