Question

A solid, horizontal cylinder of mass 12.2 kg and radius 1.00 m rotates with an angular speed of 6.00 rad/s about a fixed vertical axis through its center. A 0.250-kg piece of putty is dropped vertically onto the cylinder at a point 0.900 m from the center of rotation and sticks to the cylinder. Determine the final angular speed of the system.

Answer 1

Answer:

Final angular velocity = 5.807 rad/s

Explanation:

We know that angular momentum is given by;

L = Iω

Where;

L is angular momentum

I is moment of inertia

ω is angular velocity

Initial angular momentum is;

L_i = Iω = (1/2)mr²ω

m = 12.2kg

r = 1m

ω = 6 rad/s

Moment of inertia of cylinder = mr²/2

Moment of inertia of putty = mr²

Thus, L_i = (1/2)(12.2)(1²)(6) = 36.6 kgm²/s

Now, final angular momentum is given as;

L_f = Icyl(ω_f) + Ipuf(ω_f)

Where ;

Icyl is moment of inertia of cylinder

Ipuf is moment of inertia of puffy

ω_f is final angular velocity.

Thus,

L_f = (1/2)(m_cyl)(r²) (ω_f) + (m_puf)(r²) (ω_f)

Plugging in relevant values,

L_f = (1/2)(12.2)(1²) (ω_f) + (0.25)(0.9²)(ω_f)

L_f = (ω_f)[6.1 + 0.2025] = 6.3025(ω_f)

Now, from conservation of angular momentum, L_i = L_f

Thus, 36.6 = 6.3025(ω_f)

ω_f = 36.6/6.3025 = 5.807 rad/s