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Angular Momentum and Angular Impulse

AP Physics 1· difficulty 2/5

A 0.50 kg0.50~\text{kg} particle moves in a horizontal circle of radius 0.40 m0.40~\text{m} at 3.0 m/s3.0~\text{m/s}. Its angular momentum about the center is

  • A

    1.5 kg⋅m2/s1.5~\text{kg·m}^2/\text{s}

  • B

    2.4 kg⋅m2/s2.4~\text{kg·m}^2/\text{s}

  • C

    0.20 kg⋅m2/s0.20~\text{kg·m}^2/\text{s}

  • D

    0.60 kg⋅m2/s0.60~\text{kg·m}^2/\text{s}

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Explanation

L=mvr=(0.50)(3.0)(0.40)=0.60 kg⋅m2/sL = m v r = (0.50)(3.0)(0.40) = 0.60~\text{kg·m}^2/\text{s}.

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