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Homework Help: Calculating the velocity of a bowling shot

  1. Aug 3, 2010 #1
    1. A bowler is taking a second shot at a single pin left standing. From the top of the back swing, her arm rotates for .8 seconds at an average angular accelerations of 20.94rad/s2. The distance form the axis of roatation at her shoulder joint to the center of mass of the ball is .7m. The 7kg ball leaves her hand when her arm is just past perpendicular to the alley floor, resulting in a release angle of 10degrees counter-clockwise from the downward vertical.

    Disregarding any loss of energy due to external forces, what velocity will the 2kg pin have if the ball hits it head-on, and if the ball is slowed to 8m/s?



    2. Relevant equations

    vf=vi+at
    d=/5(vi+vf)t
    d=vit+.5at2
    vf2=vo2+2ad
    3. The attempt at a solution

    I've broken it down into angular velocity and then tangential velocity and got 11.73 m/s. That's were im stuck. I know that the tangential velocity can then be broken down into horizontal and vertical components. Not quite sure on the release angle, would it be 80degrees downward from the right horizontal, or 80 degrees downward from the left horizontal?

    The teacher gave us the final answer( the velocity of the pin after the collision is 12.43m/s) which means that the velocity of the ball before the collision is 11.55m/s. Since there is no energy lost to friction the horizontal velocity should remain the same from the initial throw to the point when it hits the pin.

    After arranging the Vt into its horizontal and vertical components I get the vertical velocity as 11.55 instead of the horizontal. I'm assuming that I'm jsut doing it wrong, but any ideas? Or ways to explain why I am?

    Thanks
     
    Last edited: Aug 3, 2010
  2. jcsd
  3. Aug 3, 2010 #2
    The vertical refers to the line directly down the bowling lane, towards the pin from the bowler, so 10° counterclockwise means it's 10° "to the left" of this vertical axis.
     
  4. Aug 3, 2010 #3
    Awesome thanks. After doing trig to find the vertical and horizontal component of the tangential velocity I get a horizontal velocity of 11.55m/s. Since there is nothing lost to friction or any resistance I can assume the velocity is 11.55 the instant before it hits the pin, correct?
     
  5. Aug 3, 2010 #4
    yes that's right. follow up with the conservation of momentum equation and you're done.
     
  6. Aug 4, 2010 #5
    If the object has a horizontal velocity of 11.55 m/s, does that mean that the bowling ball is moving along the floor at a rate of approximately 41.6 km/h?
     
  7. Aug 4, 2010 #6
    11.55 m/s is just the vertical component of the velocity. the "real" velocity is 11.73 m/s, including the horizontal component, and when converted to km/h it'd be about 42.2 km/h.
     
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