A solid cylinder of mass 12.0 kgkg and radius 0.250 mm is free to rotate without friction around its central axis. If you do 75.0 JJ of work on the cylinder to increase its angular speed, what will be its final angular speed if the cylinder starts from rest? Express your answer in radians per second.

The final angular velocity is 20rad/s

Explanation:

We are given;

mass, m = 12 kg

radius, r = 0.25 m

Work done;W = 75 J

Moment of inertia of cylinder, I = (1/2) mr²

Thus,

I = (1/2) x 12 x 0.25² = 0.375 kg.m²

Now, from work energy theorem,

Work done = Change in kinetic energy

So, W = KE_f - KE_i

Now, Initial Kinetic Energy (KE_i) = 0

Final Kinetic Energy; KE_f = (1/2)Iω²

So, KE_f = (1/2) x 0.375 x ω²

KE_f = 0.1875 ω²

Now, W = 75 J

Thus,

From, W = KE_f - KE_i, we have;

75 = 0.1875 ω² - 0

75 = 0.1875 ω²

ω² = 75/0.1875

ω² = 400

ω = √400

ω = 20 rad/s

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2