Running with a speed of 2.60m/s , a 30.0kg child heads toward the rim of a merry

Question

Running with a speed of 2.60m/s , a 30.0kg child heads toward the rim of a merry-go-round. The merry-go-round has a moment of inertia of 542kg*m^2 and a radius of 2.49m . When the child jumps onto the merry-go-round, the entire system begins to rotate.

Running with a speed of 2.60m/s , a 30.0kg child heads toward the rim of a merry-go-round. The merry-go-round has a moment of inertia of 542kg*m^2 and a radius of 2.49m . When the child jumps onto the merry-go-round, the entire system begins to rotate. Find the angle½ between the direction of motion and the outward radial direction that is required if the final angular speed of the system is to be 0.252rad/s . Answer in degrees

Explanation / Answer

angular momentum initial = angular momentum final

m vperp r = (I + mr^2) w

30*2.6*sin(theta)*2.49 = (542+30*2.6^2)*0.252

sin(theta) = 0.96638

theta = 104.9 degrees

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