So imagine a rod that receives an initial impulse acting on its tip. This impulse would cause the rod to translate and rotate about CM. Since the throw is at the tip, the rotational kinetic energy is at its max, correct? The minimum would be an impulse directly at the CM, which would cause only translation.I just want to know the ratio between the two velocities v ω when the rod is thrown by the tip.

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So imagine a rod that receives an initial impulse acting on its tip. This impulse would cause the rod to translate and rotate about CM. Since the throw is at the tip, the rotational kinetic energy is at its max, correct? The minimum would be an impulse directly at the CM, which would cause only translation.I just want to know the ratio between the two velocities       v    ω   when the rod is thrown by the tip.

Marlene Welch · Accepted Answer

So, to start, a force F perpendicular to the CM is applied at any point of the rod and this force applies an impulse into the center of mass in the value of:  ∫  F  d  t  =  p  =  m  vThis contribution goes to the translational kinetic energy:      E          k      T        =            m              v        2              2    =            p      2              2      m      Now, for the rotation contribution, the same force is applied at any given point x away from the center of mass, providing an impulse in the form of torque:  ∫  τ  d  t  =  ∫  F  x  d  t  =  p  x  =  I  ωBar moment of inertia about CM:             m              L        2              12    p  x  =            m              L        2            ω        12    ω  =            12      p      x              m              L        2            This contribution goes into the rotational kinetic energy:      E          k      R        =            I              ω        2              2    =            6              p        2                    x        2                    m              L        2            The total kinetic energy is:      E    k    =            p      2              2      m        +            6              p        2                    x        2                    m              L        2            Now, assuming that the impulse is at the tip   x  =      L    2        E    k    =            p      2              2      m        +            6              p        2                    L        2                    4      m              L        2              =            p      2              2      m        +            3              p        2                    2      m      Therefore, the rotational kinetic energy is 3 times the translational kinetic energy when the rod is thrown at the tip. Thus the translational energy is 25% the total energy and the rotational energy is 75%. Proceeding to calculate the ratio of velocities  ω  =            12      p      x              m              L        2              =            6      p              m      L        =            6      m      v              m      L        =            6      v        L        v    ω    =      L    6

Ratio between linear velocity and angular velocity in a rod thrown by the tip

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