Twin skaters approach one another as shown in the figure below and lock hands (a

Question

Twin skaters approach one another as shown in the figure below and lock hands (a) Calculate their final angular velocity, given each had an initial speed of 2.20 m/s restive to the ice. Each has a mas of as.o kg, and each has a center of mag locatad 0.820 m from their locked hands. You may approximate their maments of inertia to be that of point rsat i radius rad/s (b) Compare the initial kinetic energy and final kinetic energy O Initial kinetic energy is greater than final kinetic O Initial kinetic energy is the same as final kinetic energy O Initial kinetic energy is less than final kinetic energy Type here to search (hp

Explanation / Answer

a)

m = mass of each skater = 85 kg

v = velocity = 2.20 m/s

r = distance of center of mass from axis of rotation at locked hands = 0.83 m

I = moment of inertia of the combination = 2 m r2

w = angular velocity

Using conservation of angular momentum

mvr + mvr = I w

2 mvr = Iw

2 mvr = 2 m r2w

w = v/r = 2.20/0.83

w = 2.65 rad/s

b)

Since the two skaters gets attached to each other after collision , hence this is an example of inelastic collision. and in inelastic collision we know that there is loss of energy

hence

initial kinetic energy is greater than the final kinetic energy

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