Example 11.9 Disk and Stick Collision A 1.9 kg disk traveling at 3.1 m/s strikes

Question

Example 11.9 Disk and Stick Collision A 1.9 kg disk traveling at 3.1 m/s strikes a 1.1 stick of Before length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of figure (a). The disk strikes at the endpoint of the stick, at a distance r 1.50 m from the stick's center. Assume the collision is elastic and the disk does not deviate from its original line of motion. Find the translational speed of the disk, the translational speed of the stick, and the angular speed of the stick after the collision. The moment of inertia of the stick about its center of mass is 1.47 kg m2 SOLVE IT Conceptualize Examine figure (a) and imagine what After happens after the disk hits the stick. Figure (b) shows what you might expect: the disk continues to move at a slower speed, and the stick is in both translational and rotational motion. We assume the disk does not deviate from its original line of motion because the force exerted by the stick on the disk is parallel to the original path of the disk. Categorize Because the ice is frictionless, the disk and stick form an isolated system in terms of momentum and overhead view of a disk striking a stick in an elastic angular momentum. Ignoring the sound made in the collision. (a) Before the collision, the disk moves toward the stick. (b) The collision causes the stick collision, we also model the system as an isolated system to rotate and move to the right. in terms of energy. In addition, because the collision is assumed to be elastic, the kinetic energy of the system is Constant. Analyze First notice that we have three unknowns, so we need three equations to solve simultaneously

Explanation / Answer

the electic collision follows law of conservation of linear momentum and ke

given disc mass m1=1.9kg, moving with velocity u1=3.1m/s

collision occure with rod of mass m2=1.1kg which is at rest

according to law of conservation of momentum m1u1+m2u2=m1v1+m2v2

1.9 x3.1 +0=1.9 v1+1.1 v2

19v1+11v2=58.9 --------------------1

from this we can write v2=(58.9-19v1)/11

according to law of conservation of ke m1u12/2+m2u22/2=m1v12/2+m2v22/2

1.9 x 3.12=1.9v12+1.1v22

19v12+11v22 =182.59

sustitute v2 in the above equation we get   19v12+11((58.9-19v1)/11)2=182.59

19v12+((58.92+192v12-2 x58.9 x 19v1)/11-182.59=0   

expand and solve the above one we get 51.8v12-203.47v1+132.79=0

  v12-3.93v1+2.56=0

solve the quadradic equation we get v1=0.825m/s

substitute v1 in the above v2=(58.9-19v1)/11=(58.9-19 x 0.825)/11

=(58.9-15.675)/11=3.93m/s

the disc collide the bar at a distance r=1.5m from the center and the moment of inertia I=1.47kg m2

we know that angular momentum Iw=mvr

from this the angular velocity w=m2v2r/I =1.1 x 3.93 x 1.5/1.47

=5.823/1.47 =3.96 rad/s2

b)the gravitational force F=GMm/r2

the gravitational force decreases with increase in the height from the surface of earth

so W=GMm/(r+h)2= GMm/r2(1+h/r)2

=F/(1+h/r)2=F(1+h/r)-2=F(1-2h/r)

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