A 2.2 kg disk traveling at 2.9 m/s strikes a 1.3 kg stick of length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of figure (a). The disk strikes the stick at a distance r = 0.90 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 1.73 kg · m2. Overhead view of a disk striking a stick in an elastic collision. (a) Before the collision, the disk moves toward the stick. (b) The collision causes the stick to rotate and move to the right.

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Disk and Stick Collision
A 2.2 kg disk traveling at 2.9 m/s strikes a 1.3 kg stick of length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of figure (a). The disk strikes the stick at a distance
r = 0.90 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 1.73 kg · m2.
Overhead view of a disk striking a stick in an
elastic collision. (a) Before the collision, the disk
moves toward the stick. (b) The collision causes
the stick to rotate and move to the right.
Before
After
b
Transcribed Image Text:Disk and Stick Collision A 2.2 kg disk traveling at 2.9 m/s strikes a 1.3 kg stick of length 4.0 m that is lying flat on nearly frictionless ice as shown in the overhead view of figure (a). The disk strikes the stick at a distance r = 0.90 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 1.73 kg · m2. Overhead view of a disk striking a stick in an elastic collision. (a) Before the collision, the disk moves toward the stick. (b) The collision causes the stick to rotate and move to the right. Before After b
) + (*"s)
2v di
= 2.636
m/s
1 +
Substitute numerical values into Equation (4) to solve for w (in rad/s). (Indicate the direction with the sign of your answer.):
w = -1.7827
rad/s
Solve Equation (1) for var and substitute numerical values to solve for vur (in m/s). (Enter the magnitude.):
m
Vdi
= 2.9 m/s - 1.3 k9 (2.64 m/s)
2.2 kg
.
Vdf = V
= |1.34
V m/s
Finalize These values seem reasonable. The disk is moving more slowly
vv after the collision than it was before the collision, and the stick has a small translational speed. The table below
summarizes the initial and final values of variables for the disk and the stick, and it verifies the conservation of linear momentum, angular momentum, and kinetic energy for the isolated system. (For
angular velocity and angular momentum, indicate the direction with the signs of your answers. For the rest of the values, enter the magnitudes.)
Comparison of Values Before and After the Collision
v(m/s)
w(rad/s)
p(kg · m/s)
L(kg · m2/s)
Kerans (3)
Krot (3)
Before
Disk
2.9
6.38
5.74
9.251
Stick
Total for system
6.38
5.74
9.251
After
Disk
2.1
4.62
4.158
4.851
Stick
1.85
2.05
2.405
3.54
2.22
9.251
Total for system
7.025
7.698
7.071
9.251
Note: Linear momentum, angular momentum, and total kinetic energy of the systems are all conserved.
EXERCISE
Suppose the disk is traveling with the same velocity as in the example, but along a line that is a distance r < 2.00 m from the x-axis as shown in the figure. The disk collides elastically with the stick and
the final velocity of the disk is found to be 1.34 m/s along its original line of motion. Determine the distance r (in m).
Hint
0.821
Use the conservation of linear and angular momentum which applies in general, and the conservation of kinetic energy that applies specifically for an elastic collision, to find the relations between initial
and final linear and rotational velocities. m
Transcribed Image Text:) + (*"s) 2v di = 2.636 m/s 1 + Substitute numerical values into Equation (4) to solve for w (in rad/s). (Indicate the direction with the sign of your answer.): w = -1.7827 rad/s Solve Equation (1) for var and substitute numerical values to solve for vur (in m/s). (Enter the magnitude.): m Vdi = 2.9 m/s - 1.3 k9 (2.64 m/s) 2.2 kg . Vdf = V = |1.34 V m/s Finalize These values seem reasonable. The disk is moving more slowly vv after the collision than it was before the collision, and the stick has a small translational speed. The table below summarizes the initial and final values of variables for the disk and the stick, and it verifies the conservation of linear momentum, angular momentum, and kinetic energy for the isolated system. (For angular velocity and angular momentum, indicate the direction with the signs of your answers. For the rest of the values, enter the magnitudes.) Comparison of Values Before and After the Collision v(m/s) w(rad/s) p(kg · m/s) L(kg · m2/s) Kerans (3) Krot (3) Before Disk 2.9 6.38 5.74 9.251 Stick Total for system 6.38 5.74 9.251 After Disk 2.1 4.62 4.158 4.851 Stick 1.85 2.05 2.405 3.54 2.22 9.251 Total for system 7.025 7.698 7.071 9.251 Note: Linear momentum, angular momentum, and total kinetic energy of the systems are all conserved. EXERCISE Suppose the disk is traveling with the same velocity as in the example, but along a line that is a distance r < 2.00 m from the x-axis as shown in the figure. The disk collides elastically with the stick and the final velocity of the disk is found to be 1.34 m/s along its original line of motion. Determine the distance r (in m). Hint 0.821 Use the conservation of linear and angular momentum which applies in general, and the conservation of kinetic energy that applies specifically for an elastic collision, to find the relations between initial and final linear and rotational velocities. m
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