Question

A rod rotates in a horizontal plane with angular speed on into the page) the instant it is struck by a bullet travelling as shown. The bullet passes through the end of the rod, leaving a hole in the rod, and continues in the same direction. a) Give the equation needed to find the angular velocity of the rod just after the collision (leave it in terms of the symbols given below for part a). b) (You should be able to solve part b without doing part a, they are independent) The rod is rotating with angular speed 5 rad/s just after the collision. If the axis of rotation has friction that produces a constant torque of 0.15 Nm, how many revolutions does it rotate through before coming to rest? (Use mg = 0.4 kg and L = 1.2 m for part b only] 0. Axis of Rotation 1/3 L Given information (part a): Rod: Mass = mr Length = L Initial angular speed = 0, (into the page) Bullet: Mass = mg Initial speed = V Final speed = v/2 2/3 L 70° Top Down View (as viewed from above)

Answer 1

wir A.0.R. mass of rod=MR MB alt v mB oh (a) Impulse on bullet more fFdt = me -more SF dt = more - ow (final angular speed) alr angular Impulse on rod X21 X2L dt = I w-I wo moment of Inertia oftod about opio of rotation -ff sino

I = me² + 12 2 I= ma (+)² mee – 24 samo ffot e mall [w-ws] from (1) & (2) equation. - 24 Smo may MR 2² [w-wo 3 W=Woh Wo-3mB u sino MRL o= 70 sin 70° = 0.94 w= Wo- 3 moux (0.94) MRL W = Wo - 2.82 more MRL (6) wasrad/5 L = 1.2m MR = 0.4kg. Constant friction torque Tf = 0.15 N-m this torque produces constant angular See retardation (&) If = I a

TH MR LL lgtf da MRL? 다. 2 = gx 0.15 0.4% (1.2)? x= 2.34 radle angle rotated by rod before coming to rest 80 is o w?= w?-200 O = (5)2– 2x2.34 0 = 5.342 rad. No of revolution () = 21 S.342 N = 2X X N = 0.85