Tutorial Exercise The position vector of a particle of mass 2.55 kg as a functio

Question

Tutorial Exercise The position vector of a particle of mass 2.55 kg as a function of time is given by r - (6.00 î 4.15 t), where r is in meters and t is in seconds. Determine the angular momentum of the particle about the origin as a function of time Part 1 of 4 - Conceptualize Think of plotting the particle's trajectory. It moves parallel to the y axis along the line x = 6.00 m, with speed 4.15 m/s. Its angular momentum as it crosses the x axis will be 2.55(6.00) (4.15)- 63.5 units pointing out of the plane parallel to the z axis Part 2 of 4 Categorize We will find the particle's velocity by differentiating its position, and then use the equation L = m r· Part 3 of 4 Analyze The velocity of the particle is given by the following 6im+ 4.15 4.15 tj m4.15 4.15 j m/s dt dt Part 4 of 4-Analyze The angular momentum is given by so we have the following L=( 2.55 kg)6.00 m + 4.15 tj m) 4.15 m/s Because we know that j = k and j j = 0, we can simplify i-(63.495 kg . m2/s)i.j] + [( 4.15 L = ( 63.495 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. t kg - m2/s j x j to obtain L-( 63.5 k) kg m2/s Since the t term has been eliminated, L is constant in time SubmitSkip (you cannot come back)

Explanation / Answer

r = 6 i + 4.15 t j

v = dr/dt = 0i + 4.15 j

L = m ( r x v)

L = 2.55 [ (6 i + 4.15t j) x (4.15j)]

L = 2.55[ 24.9 k + 0 ]

L = (63.5) (i x j) + (17.22 t) (j x j)

L = 63.54 k + 0

Ans: L = 63.5 k

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