A rod of mass M = 154 g and length L = 35 cm can rotate about a hinge at its left end and is initially at rest. A putty ball of mass m = 11 g, moving with speed V = 9 m/s, strikes the rod at angle θ = 29° a distance D = L/3 from the end and sticks to the rod after the collision. Calculate the rotational kinetic energy, in joules, of the system after the collision.

moment of inertia of the rod = 1/3 mL², m is mass and L is length of rod.

1/3 x. 154 x. 35²

=.00629

moment of inertia of putty about the axis of rotation

= m d², m is mass of putty and d is distance fro axis

=.011 x (.35 / 3) ²

=.00015

Total moment of inertia I =.00644 kgm²

angular momentum of putty about the axis of rotation

= mvRsinθ

m is mass, v is velocity, R is distance where it strikes the rod and θ is angle with the rod at which putty strikes

=.011 x 9 x. 35 / 3 x sin 29

=.0056

Applying conservation of angular momentum

angular momentum of putty = angular momentum of system after of collision

.0056 =.00644 ω where ω is angular velocity of the rod after collision

ω =.87 rad / s.

Rotational energy

= 1/2 I ω²

I is total moment of inertia

=.5 x. 00644 x. 87²

= 2.44 x 10⁻³ J.