A uniform rod of length L rests on a frictionless horizontal surface. The rod pi

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.

What is the final angular speed of the rod?

Express your answer in terms of the variables v and L.

Answer is w=6v/19L

What is the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision? Answer does not include variables!

Explanation / Answer

here,

mass of the rod is m

mass of the bullet is m/4

let the length of rod be l

let the final angular speed of the rod is w

using conservation of angular momentum

(m/4) * v * (l/2) = ( (m/4)*(l/2)^2 + m*l^2/3 ) * w

w = ((m/4) * v * (l/2) ) /((m/4)*(l/2)^2 + m*l^2/3 )

w = 6v/19L

the angular speed of the rod is 6v/19L

the ratio of the kinetic energy of the system after the collision to the kinetic energy of the bullet before the collision , KE

KE = KEb/KEa

KE = 0.5 * m * v^2 /( 0.5 * I*w^2)

KE = 0.5 * m * v^2 /( 0.5 * ((m/4)*(l/2)^2 + m*l^2/3 )*w^2)

KE = 0.5 * m * v^2 /( 0.5 * ( 0.146*m*l^2 )*(0.87 * v)^2)

KE = 8

the kinetic energy ratio is 8

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.