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3 years ago

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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.

1 answer:

dybincka [34]3 years ago

5 0

Answer:

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

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