A projectile of mass msubp is traveling at a constant velocityvsu0 toward a stat

Question

A projectile of mass msubp is traveling at a constant velocityvsu0 toward a stationary disk of mass M and radius R that is freeto rotate its axis (figure 10-54). Before impact, the projectile istraveling along a line displaced a distance b below the axis. Theprojectile strikes the disk and sticks to point B. Model theprojectile as a point mass.
A.) Before impact what is the total angular momentum Lsub0 of thedisk-projectile system about the axis? Answer the followingquestions in terms of the symbols given at the strat of theproblem. B.) What is the angular speed of the disk-projectilesystem just after the impact? C.) What is the kinetic energy of the disk-projectile systemafter impact? D.) How much mechanical energy is lost in thiscollision? A projectile of mass msubp is traveling at a constant velocityvsu0 toward a stationary disk of mass M and radius R that is freeto rotate its axis (figure 10-54). Before impact, the projectile istraveling along a line displaced a distance b below the axis. Theprojectile strikes the disk and sticks to point B. Model theprojectile as a point mass.
A.) Before impact what is the total angular momentum Lsub0 of thedisk-projectile system about the axis? Answer the followingquestions in terms of the symbols given at the strat of theproblem. B.) What is the angular speed of the disk-projectilesystem just after the impact? C.) What is the kinetic energy of the disk-projectile systemafter impact? D.) How much mechanical energy is lost in thiscollision?

Explanation / Answer

The angular momentum is conserved in this perfectly inelasticcollision ,because the net external torque acting on the system iszero. a) The total angular momentum of the disk & the projectilejust before impact is     L0 =mpvob b) The angular momentum of the disk-projectile system just afterthe impact is        =L0/I But the moment of inertia of the disk plus the projectile is I= 1/2*MR2+mpb2 Thus     =2mpv0b/MR2+2mpb2 c) The kinetic energy of the system after impact in termsof its angular momentum is        Kf =L2/2I             =(mpvob)2/2(1/2*MR2+mpb2)             =(mpvob)2/MR2+2mpb2 d) The mechanical energy is lost in this collision is thedifference between the initial & final kinetic energies, whichis E = Ki-Kf         =1/2*mpvo2 -(mpvob)2/MR2+2mpb2           =1/2*mpvo2( 1-mpb2/MR2+2mpb2)           =1/2*mpvo2( 1-mpb2/MR2+2mpb2)

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