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a uniform rod of length l rests on a frictionless horizontal surface. the rod pivots about a fixed frictionless axis at one end. the rod is initially at rest. a bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. the mass of the bullet is one-fourth the mass of the rod.

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get a uniform rod of length l rests on a frictionless horizontal surface. the rod pivots about a fixed frictionless axis at one end. the rod is initially at rest. a bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. the mass of the bullet is one-fourth the mass of the rod. from EN Bilgi.

A uniform rod of length L rests on a frictionless horizontal

Find step-by-step Physics solutions and your answer to the following textbook question: A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?.

Question

A uniform rod of length L rests on a frictionless horizontal surface. The rod pivots about a fixed frictionless axis at one end. The rod is initially at rest. A bullet traveling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its center and becomes embedded in it. The mass of the bullet is one-fourth the mass of the rod. (a) What is the final angular speed of the rod? (b) What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision?

Explanation

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a) To solve the first part of this problem, we have to assume that since external forces which are acting on the givens system do not produce any torque, the Law of conservation of angular momentum can be applied.

The Law of conservation of angular momentum states that initial and final angular momentum will be equal, total angular momentum will remain constant, whenever there is no torque produced by the external forces. Therefore, it can be expressed by the following equation:

L_{1} = L_{2} L 1 ​ =L 2 ​ Where: L_{1} L 1 ​

- initial angular momentum

L_{2} L 2 ​

- final angular momentum

Also, we will be applying an equation that determines angular momentum:

L = I \omega L=Iω Where: I

I - moment of inertia

\omega ω - angular speed

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Solved A uniform rod of length L rests on a frictionless

Answer to Solved A uniform rod of length L rests on a frictionless

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A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one

Click here👆to get an answer to your question ✍️ A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one - sixth the mass of the rod. What is the final angular velocity of the rod ?

Question

A uniform rod of length L rests on a frictionless horizontal surface. The rod is pivoted about a fixed frictionless axis at one end. The rod is initially at rest. A bullet travelling parallel to the horizontal surface and perpendicular to the rod with speed v strikes the rod at its centre and becomes embedded in it. The mass of the bullet is one-sixth the mass of the rod. What is the final angular velocity of the rod ?

Aω=

9L v ​

Bω=

9L 2v ​

Cω=

9L 3v ​

Dω=

9L 5v ​ Hard Open in App Solution Verified by Toppr

Correct option is B)

Angular momentum about O remains constant just before and just after collision

∴L i ​ =L f ​ or ( 6 m ​ v 2 L ​ )=Iω =[ 3 mL 2 ​ + 6 m ​ 4 L 2 ​ ]ω Solving, we get 12 mvL ​ = 24 9mL 2 ​ ω= 9L 2v ​

Video Explanation