Question

A uniform solid disk made of wood is horizontal and rotates freely about a vertical axle at its center. The disk has radius 0.600 m and mass 1.60 kg and is initially at rest. A bullet with mass 0.0200 kg is fired horizontally at the disk, strikes the rim of the disk at a point perpendicular to the radius of the disk, and becomes embedded in its rim, a distance of 0.600 m from the axle. After being struck by the bullet, the disk rotates at 4.00 rad/s. What is the horizontal velocity of the bullet just before it strikes the disk?

Answer 1

Answer:

v = 98.4 m/s

Explanation:

As we know that the total angular momentum of the system is conserved

so here we will have

$L_{bullet} + L_{disk} = constant$

$MvR = (\frac{1}{2}mR^2 + MR^2)\omega$

now plug in all data

$0.02 v(0.600) = (\frac{1}{2}(1.60)(0.600)^2 + 0.02(0.600)^2)4$

$0.012 v = 1.18$

$v = \frac{1.18}{0.012}$

$v = 98.4 m/s$