AP Physics 1 · Topic 5.5

Rotational Equilibrium and Newton's First Law in Rotational Form Practice

Part of Torque and Rotational Dynamics.(TOP-5.E)

Practice questions

21

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Sample questions

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  1. Sample 1difficulty 1/5

    30 kg 30 kg

    Two equal masses sit at equal distances from the pivot of a uniform seesaw. The system is

    • A

      Rotating clockwise

    • B

      In rotational equilibrium

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    • C

      Rotating counterclockwise

    • D

      About to tip

    Why

    Equal torques on both sides cancel — net torque zero, so rotational equilibrium.

  2. Sample 2difficulty 2/5

    25 kg 40 kg 1.6 m d

    A 25 kg25~\text{kg} child sits 1.6 m1.6~\text{m} from the pivot. To balance, a 40 kg40~\text{kg} child must sit at distance

    • A

      1.5 m1.5~\text{m}

    • B

      2.5 m2.5~\text{m}

    • C

      1.0 m1.0~\text{m}

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    • D

      0.5 m0.5~\text{m}

    Why

    251.6=40dd=40/40=1.0 m25 \cdot 1.6 = 40 \cdot d \Rightarrow d = 40/40 = 1.0~\text{m}.

  3. Sample 3difficulty 2/5

    A student applies a force at the rim of a wheel of radius 0.40 m0.40~\text{m} to produce 20 N⋅m20~\text{N·m} clockwise torque. A second force of 80 N80~\text{N} is applied tangentially at the rim counterclockwise. Is the wheel in rotational equilibrium?

    • A

      No — the counterclockwise torque wins.

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    • B

      No — the clockwise torque wins.

    • C

      Cannot tell without the wheel's mass.

    • D

      Yes — both torques are equal.

    Why

    CCW torque: (0.40)(80)=32 N⋅m(0.40)(80) = 32~\text{N·m} > 20 N⋅m20~\text{N·m} CW. Net torque is counterclockwise, so not in equilibrium.

  4. Sample 4difficulty 2/5

    40 kg m 1.5 m 2.0 m

    A 40 kg40~\text{kg} child sits 1.5 m1.5~\text{m} to the left of a seesaw's fulcrum. To balance, what mass must be placed 2.0 m2.0~\text{m} to the right?

    • A

      30 kg30~\text{kg}

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    • B

      53 kg53~\text{kg}

    • C

      40 kg40~\text{kg}

    • D

      20 kg20~\text{kg}

    Why

    Balance: m1gd1=m2gd2m2=(40)(1.5)/2.0=30 kgm_1 g d_1 = m_2 g d_2 \Rightarrow m_2 = (40)(1.5)/2.0 = 30~\text{kg}.

  5. Sample 5difficulty 3/5

    For the previous setup, scale <strong>A</strong> reads

    • A

      125 N125~\text{N}

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    • B

      50 N50~\text{N}

    • C

      175 N175~\text{N}

    • D

      100 N100~\text{N}

    Why

    Total weight: 300 N300~\text{N}. FA=300175=125 NF_A = 300 - 175 = 125~\text{N}.