Answer:
The final angular speed is 7.71 rad/s
Explanation:
Given
Cylinder mass, M = 10.6 kg
Cylinder radius, R = 1.00 m
Angular speed, w = 8.00 rad/s.
Mass of putty, m = 0.250-kg
Radius, r = 0.900 m
First, we set up an expression for the initial and final angular momentum of the system.
The moment of inertia of the cylinder is given as I = ½MR²
While the moment of inertia if the putty is mr².
Initial Momentum of the system = Initial momentum of the cylinder =
Li = Iw --- Substitute ½MR² for I
Li = ½MR²w
By
Substituton
Li = ½ * 10.6 * 1² * 8
Li = 42.4kgm²/s
Calculating the final momentum of the system.
First we calculate the final momentum of the cylinder
Li = Iw --- Substitute ½MR² for I
Li = ½MR²wf where wf = final angular speed
By
Substituton
Li = ½ * 10.6 * 1² wf
Li = 5.3w kgm²/s
Then we calculate the final momentum of the putty
Final Momentum of the putty =
L2 = Iwf --- Substitute mr² for I;
L2 = mr²wf --- By Substituton
L2 = 0.25 * 0.9² * wf
L2 = 0.2025wf kgm²/s
Final momentum = Li + L2
Lf = (5.3wf + 0.2025wf) kgm²/s
Lf = 5.5025wf kgm²/s
By conservation of momentum
Li = Lf
Where Li = 42.4kgm²/s and Lf = 5.5025wf kgm²/s
So, we have
5.5025wf kgm²/s = 42.4kgm²/s --- make wf the subject of formula
wf = 42.4/5.5025
wf = 7.71 rad/s
Hence, the final angular speed is 7.71 rad/s
