Newton’s First & Second Law in Rotational Form (College Board AP® Physics 1: Algebra-Based): Flashcards

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  • Define rotational equilibrium.

    Rotational equilibrium is a configuration of torques such that the net torque exerted on the system is zero.

    \sum \tau_{i} = 0

  • What is the angular acceleration of a system in rotational equilibrium?

    It is zero, because the net torque exerted on the system is zero.

  • State the principle of torques.

    For a system in rotational equilibrium, the sum of the clockwise torques equals the sum of the counterclockwise torques.

  • How is a balanced beam problem solved?

    • Identify the location of the pivot and calculate the torque due to each force about it

    • Sum the clockwise and counterclockwise torques separately

    • Use the principle of torques to calculate the unknown quantity

  • Rotational equilibrium is analogous to .......... equilibrium.

    Rotational equilibrium is analogous to translational equilibrium.

  • True or False?

    A system in rotational equilibrium must be at rest.

    False.

    The system may also rotate with a constant angular velocity, because the net torque is zero and there is no angular acceleration.

  • State Newton's first law in rotational form.

    If the net torque exerted on a system is zero, the angular velocity of that system remains constant.

  • If the net torque exerted on a system is zero, the system is said to be in .......... equilibrium.

    If the net torque exerted on a system is zero, the system is said to be in rotational equilibrium.

  • True or False?

    A non-rotating system cannot have a constant angular velocity.

    False.

    A constant angular velocity can be an angular velocity of zero, when the system is not rotating.

  • What happens to the angular velocity of a rigid system if the torques exerted on it are not balanced?

    The angular velocity changes, because a net torque produces an angular acceleration.

  • What determines the direction of the angular acceleration of a rigid system?

    The direction of the net torque.

  • How do you find the net torque from clockwise and counterclockwise torques?

    Subtract one total from the other:

    \sum \tau = \tau_{\text{clockwise}} - \tau_{\text{counterclockwise}}

    The net torque acts in the direction of the larger total.

  • State Newton's second law in rotational form.

    The angular acceleration of a rigid system is directly proportional to the net torque exerted on it and is in the same direction.

  • State the equation for the angular acceleration of a system.

    \alpha_{\text{sys}} = \frac{\sum \tau}{I_{\text{sys}}} = \frac{\tau_{\text{net}}}{I_{\text{sys}}}

    • \alpha_{\text{sys}} = angular acceleration of the system (rad/s2)

    • \tau_{\text{net}} = net torque exerted on the system (N·m)

    • I_{\text{sys}} = rotational inertia of the system (kg·m2)

  • In a pulley-mass system, what exerts the torque on the pulley?

    The tension in the string, acting at the radius of the pulley.

  • What is the rotational analog of mass in Newton's second law?

    Rotational inertia, I.

    F_{\text{net}} = ma becomes \tau_{\text{net}} = I\alpha

  • Unbalanced torques produce .......... acceleration.

    Unbalanced torques produce angular acceleration.

  • True or False?

    A larger rotational inertia gives a larger angular acceleration for the same net torque.

    False.

    Angular acceleration is inversely proportional to rotational inertia, so a larger rotational inertia gives a smaller angular acceleration.

  • State the relationships between tangential and angular motion.

    v_{T} = r\omega

    a_{T} = r\alpha

    • v_{T} = tangential velocity (m/s)

    • a_{T} = tangential acceleration (m/s2)

    • r = distance from the axis of rotation (m)

    • \omega = angular velocity (rad/s)

    • \alpha = angular acceleration (rad/s2)

  • Why may linear and rotational quantities need to be calculated separately for a pulley-mass system?

    Torques and linear forces act independently on the system.

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