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Photo courtesy Wikimedia Commons, by user Fir0002

When a ball is started out with a purely sliding motion on a slick surface like a bowling alley, it will gradually begin to rotate until eventually (if the lane is long enough) it is rolling without slipping. Suppose a bowling ball which is effectively a solid sphere with a diameter of 22 cm and a mass of 6.0 kg is started down a perfectly level alley with a coefficient of kinetic friction equal to 0.10. The ball is initially sliding with no rotation and is moving at a rate of 2.8 m/s. How fast will the center of mass of the ball be moving along the alley when it is rolling without slipping?

Solution

We will solve the problem via three different methods.

    Method 1

    System:

    Interactions:

    Model:

    Approach:

    Diagrammatic Representation

    We begin with a free body diagram. Since this problem will deal with rotation, we must also select an axis of rotation.

    In this case, the most direct (but probably not the most obvious) method of solution is to consider the angular momentum about an axis fixed at some point on the alley surface. For simplicity, we select the point at which the ball is released. By taking a point on the alley surface, we have guaranteed that there is zero net torque acting on the system (the ball) about this axis. Gravity and the normal force each create a nonzero torque, but they balance each other perfectly because they share the same moment arm and they have the same magnitude. The friction force produces no torque because it has zero lever arm.

    If it is not apparent to you that the force of gravity and the normal force acting on the ball have equal magnitudes, you should write Newton's 2nd Law for the y-direction.

    Mathematical Representation

    The [angular momentum (one-dimensional)] of the ball about the axis we have chosen can be broken into two parts by considering the contribution of the translation of the ball's center of mass and the rotation of the ball about its center of mass.

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    \begin

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    [ L = L_

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    + L_

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    = mv + I\omega ] \end

    Since there is zero net torque acting on the ball, the angular momentum is conserved and we can write:

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    \begin

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    [ mv_

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    = mv_

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    ] \end

    We know that ωi = 0, and we also know that the ball ends up rolling without slipping. Thus, we know that:

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    \begin

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    [ \omega_

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    ]\end

    Thus, we have:

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    \begin

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    [ v_

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    = \frac{v_{i}}{1+\frac

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    {\displaystyle mR^

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    }} = \frac{v_{i}}{1+\frac

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    {\displaystyle 5}} = \frac

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    v_

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    = \mbox

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    ]\end

    Method 2

    System:

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    Approach:

    Method 3

    System:

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