Question

A disk-shaped merry-go-round of radius 2.83 m and mass 185 kg rotates freely with an angular speed of 0.701 rev/s . A 63.4 kg person running tangential to the rim of the merry-go-round at 3.51 m/s jumps onto its rim and holds on. Before jumping on the merry-go-round, the person was moving in the same direction as the merry-go-round's rim. Part A What is the final angular speed of the merry-go-round

Answer 1

Answer:

The final angular speed of the merry-go-round is $3.118\,\frac{rad}{s}$ $\left(0.496\,\frac{rev}{s} \right)$.

Explanation:

Given the absence of external forces, the final angular speed of the merry-go-round can be determined with the resource of the Principle of Angular Momentum Conservation, which is described in this case as:

$I_{g, m} \cdot \omega_{o,m} + I_{g, p}\cdot \omega_{o,p} = (I_{g,m} + I_{g, p})\cdot \omega_{f}$

Where:

$I_{g,m}$ - Moment of inertia of the merry-go-round with respect to its axis of rotation, measured in $kg\cdot m^{2}$.

$I_{g,p}$ - Moment of inertia of the person with respect to the axis of rotation of the merry-go-round, measured in $kg\cdot m^{2}$.

$\omega_{o, m}$ - Initial angular speed of the merry-go-round with respect to its axis of rotation, measured in radians per second.

$\omega_{o,p}$ - Initial angular speed of the merry-go-round with respect to the axis of rotation of the merry-go-round, measured in radians per second.

$\omega_{f}$ - Final angular speed of the merry-go-round-person system, measured in radians per second.

The final angular speed is cleared:

$\omega_{f} = \frac{I_{g,m}\cdot \omega_{o,m}+I_{g,p}\cdot \omega_{o,p}}{I_{g,m}+I_{g,p}}$

Merry-go-round is modelled as uniform disk-like rigid body, whereas the person can be modelled as a particle. The expressions for their moments of inertia are, respectively:

Merry-go-round

$I_{g,m} = \frac{1}{2}\cdot M \cdot R^{2}$

Where:

$M$ - The mass of the merry-go-round, measured in kilograms.

$R$ - Radius of the merry-go-round, measured in meters.

Person

$I_{g,p} = m\cdot r^{2}$

Where:

$m$ - The mass of the person, measured in kilograms.

$r$ - Distance of the person with respect to the axis of rotation of the merry-go-round, measured in meters.

If $M = 185\,kg$, $m = 63.4\,kg$, $R = r = 2.83\,m$, the moments of inertia are, respectively:

$I_{g,m} = \frac{1}{2}\cdot (185\,kg)\cdot (2.83\,m)^{2}$

$I_{g,m} = 740.823\,kg\cdot m^{2}$

$I_{g,p} = (63.4\,kg)\cdot (2.83\,m)^{2}$

$I_{g,p} = 507.764\,kg\cdot m^{2}$

The angular speed experimented by the person with respect to the axis of rotation of the merry-go-round is:

$\omega_{o,p} = \frac{v_{p}}{r}$

$\omega_{o,p} = \frac{3.51\,\frac{m}{s} }{2.83\,m}$

$\omega_{o,p} = 1.240\,\frac{rad}{s}$

Given that $I_{g,m} = 740.823\,kg\cdot m^{2}$, $I_{g,p} = 507.764\,kg\cdot m^{2}$, $\omega_{o,m} = 4.405\,\frac{rad}{s}$ and $\omega_{o,p} = 1.240\,\frac{rad}{s}$, the final angular speed of the merry-go-round is:

$\omega_{f} = \frac{(740.823\,kg\cdot m^{2})\cdot \left(4.405\,\frac{rad}{s} \right)+(507.764\,kg\cdot m^{2})\cdot \left(1.240\,\frac{rad}{s} \right)}{740.823\,kg\cdot m^{2}+507.764\,kg\cdot m^{2}}$

$\omega_{f} = 3.118\,\frac{rad}{s}$

$\omega_{f} = 0.496\,\frac{rad}{s}$

The final angular speed of the merry-go-round is $3.118\,\frac{rad}{s}$ $\left(0.496\,\frac{rev}{s} \right)$.