One ice skater of mass

*m*moves with speed 2*v*, while another of the same mass*m*moves with speed*v*toward the left, as shown in Figure I. Their paths are separated by a distance*b*. At*t*= 0, when they are both at*x*= 0, they grasp a pole of length*b*and negligible mass. For*t*> 0, consider the system as a rigid body of two masses*m*separated by distance*b*, as shown in Figure II. Which of the following is the correct formula for the motion after*t*= 0 of the skater initially at*y*=*b*/2?A. | x = 2vt |
y = b/2 |

B. | x = vt + 0.5b sin (3vt/b) |
y = 0.5b cos (3vt/b) |

C. | x = 0.5vt + 0.5b sin (3vt/b) |
y = 0.5b cos (3vt/b) |

D. | x = vt + 0.5b sin (6vt/b) |
y = 0.5b cos (6vt/b) |

E. |
x = 0.5vt + 0.5b sin (6vt/b) |
y = 0.5b cos (6vt/b) |

**(GR9277 #78)**

**Solution:**

*x*=

*x*

_{translational}+

*x*

_{rotational}

To find

*x*

_{trans}:

Conservation of linear momentum:

*m*₁

*v*₁ +

*m*₂

*v*₂ = (

*m*₁ +

*m*₂)

*v*'

*m*(2

*v*) +

*m*(−

*v*) = 2

*m*

*v*'

→

*v*' = 0.5

*v*

→

_{ }x_{trans}=

*v*'

*t*= 0.5

*vt*

→ A, B and E are FALSE.

Check answer C.

*x*= 0.5

*vt*+ 0.5

*b*sin (3

*vt*/

*b*)

*→*x_{rot}=

*a*sin

*ωt =*0.5

*b*sin (3

*vt*/

*b*)

*→**ω =*3

*v*/

*b*

To check if

*ω =*3

*v*/

*b*:

At

*t*= 0,

*L =*

*L*₁ +

*L*₂

*= r*₁

*m*₁

*v*₁

*+*

*r*₂

*m*₂

*v*₂ =

^{1}/

_{2}

*bm*2

*v*

*+*

^{1}/

_{2}

*b*

*m*

*v =*

^{3}/

_{2}

*bm*

*v*

At

*t*> 0,

*L*' =

*Iω*

Moment inertia of the system:

*I*= ½

*mb*²

*→ L*' = ½

*²*

*mb**ω*

Conservation of angular momentum:

*L = L*'

*→*

^{ }^{3}/

_{2}

*bm*

*v =*

^{1}/

_{2}

*²*

*mb**ω*→

^{ }

*ω =*3

*v/b*

**Answer:**C