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

The final angular speed of the merry-go-round is
`3.118\,(rad)/(s)`
`\left(0.496\,(rev)/(s) \right)`.

Step-by-step 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) = (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) = (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) = (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) = (v_(p))/(r)`

`\omega_(o,p) = (3.51\,(m)/(s) )/(2.83\,m)`

`\omega_(o,p) = 1.240\,(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\,(rad)/(s)` and
`\omega_(o,p) = 1.240\,(rad)/(s)`, the final angular speed of the merry-go-round is:

`\omega_(f) = ((740.823\,kg\cdot m^(2))\cdot \left(4.405\,(rad)/(s) \right)+(507.764\,kg\cdot m^(2))\cdot \left(1.240\,(rad)/(s) \right))/(740.823\,kg\cdot m^(2)+507.764\,kg\cdot m^(2))`

`\omega_(f) = 3.118\,(rad)/(s)`

`\omega_(f) = 0.496\,(rad)/(s)`

The final angular speed of the merry-go-round is
`3.118\,(rad)/(s)`
`\left(0.496\,(rev)/(s) \right)`.