A 4.70 kg ball is dropped from a height of 13.0 m above one end of a uniform bar that pivots at its center. The bar has mass 6.50 kg and is 4.00 m in length. At the other end of the bar sits another 5.40 kg ball, unattached to the bar. The dropped ball sticks to the bar after the collision.
How high will the other ball go after the collision?
How high will the other ball go after the collision?
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this problem involves conservation of angular momentum
just before impact, the dropping ball has angular momentum with respect to the pivot of
m v r = 4.7kg x 16m/s x 2m
16m/s is the speed acquired dropping through 13m, and 2m is the perpendicular distance from the pivot, thus, there is an initial angular momentum of 150 kgm^2/s
the rod with the masses attached will begin to rotate, and the entire assembly will have an angular momentum of 150kgm^2/s
the angular momentum of the assembly will equal I w where I is the moment of inertia and w the angular velocity
we want to find the angular velocity to determine the launch velocity of the 5.4kg ball
first, find the moment of inertia
the moment of inertia will be equal to the sum of the moment of inertia of a thin rod pivoted through its middle, plus the moments of inertia of two discrete masses 2m from the pivot
this gives us a total moment of inertia of:
1/12 (6.5kg)(4m)^2 + 5.4kg x (2m)^2+ 4.7kg x (2m)^2 = 49kgm^2
this means that 150kgm^2/s = 49kgm^2 x w
w = 3.06rad/s and this is the angular velocity of the board
the linear velocity of the end of board will be v = w x 2m = 6.12m/s
therefore, the mass at the end will be launched with a linear velocity of 6.12m/s, and will attain a height of v^2/2g = 1.91m
just before impact, the dropping ball has angular momentum with respect to the pivot of
m v r = 4.7kg x 16m/s x 2m
16m/s is the speed acquired dropping through 13m, and 2m is the perpendicular distance from the pivot, thus, there is an initial angular momentum of 150 kgm^2/s
the rod with the masses attached will begin to rotate, and the entire assembly will have an angular momentum of 150kgm^2/s
the angular momentum of the assembly will equal I w where I is the moment of inertia and w the angular velocity
we want to find the angular velocity to determine the launch velocity of the 5.4kg ball
first, find the moment of inertia
the moment of inertia will be equal to the sum of the moment of inertia of a thin rod pivoted through its middle, plus the moments of inertia of two discrete masses 2m from the pivot
this gives us a total moment of inertia of:
1/12 (6.5kg)(4m)^2 + 5.4kg x (2m)^2+ 4.7kg x (2m)^2 = 49kgm^2
this means that 150kgm^2/s = 49kgm^2 x w
w = 3.06rad/s and this is the angular velocity of the board
the linear velocity of the end of board will be v = w x 2m = 6.12m/s
therefore, the mass at the end will be launched with a linear velocity of 6.12m/s, and will attain a height of v^2/2g = 1.91m