Question

I I) A 4.2-m-diameter merry-g o-round is rotating freely with a n angular velocity of 0.80 r ad/s. Its to tal m om ent of ine rtia is 1760 kg • m2. Four people standing on the ground, each of mass 65 kg, suddenl y step onto the edge of the merry-go-round. What is the angular velocity of th e merry-go-round now? W ha t if the people were o n it initially and the n jum ped off in a radial direction (relative to the merry-go -round)?

Answer 1

Answer:

a) $\omega = 0.432\,\frac{rad}{s}$, b) A portion of the initial total energy is removed from system by the jump of the people and angular speed does not change. In other word, merry-go-round rotates slower.

Explanation:

a) Let assume that each person can be modelled as particles. The system is modelled after the Principle of Angular Momentum Conservation:

$(1760\,kg\cdot m^{2})\cdot (0.80\,\frac{rad}{s}) = [1760\,kg\cdot m^{2} + 4\cdot (65\,kg)\cdot (2.4\,m)^{2}]\cdot \omega$

The angular speed is:

$\omega = 0.432\,\frac{rad}{s}$

b) A portion of the initial total energy is removed from system by the jump of the people and angular speed does not change. In other word, merry-go-round rotates slower. According to the Principle of Energy Conservation, the phenomenon is observed:

$K_{A} = K_{B,m} + 4\cdot K_{B,p}$

$\frac{1}{2}\cdot (I_{m}+4\cdot I_{p})\cdot \omega^{2} = \frac{1}{2}\cdot I_{m}\cdot \omega^{2} + 2\cdot m\cdot v^{2}$

$2\cdot I_{p}\cdot \omega^{2} = 2 \cdot m \cdot v^{2}$

$m\cdot R^{2}\cdot \omega^{2} = m\cdot v^{2}$

$v^{2} = R^{2}\cdot \omega^{2}$

$v = R\cdot \omega$