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A projectile of mass 0.2 kg and an initial velocity of 50 m/s collides with the end of a blade attached to a turbine. The rotational inertia of the turbine is 12.5 kg⋅m2 . Assume the loss of energy of the projectile in the collision is completely transferred to the blades, causing them to spin. If the final rebound velocity of the projectile after hitting a turbine blade is −25 m/s , which of the following is most nearly the rotational velocity of the turbine after the collision?

User Kebs
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1 Answer

9 votes

Answer:

5.47 rad/s

Step-by-step explanation:

We can solve this problem using the principle of conservation of energy and angular momentum.

First, let's find the initial kinetic energy of the projectile:

K1 = (1/2) * m * v1^2

= (1/2) * 0.2 kg * (50 m/s)^2

= 250 J

where v1 is the initial velocity of the projectile.

Next, let's find the final kinetic energy of the projectile:

K2 = (1/2) * m * v2^2

= (1/2) * 0.2 kg * (-25 m/s)^2

= 67.5 J

where v2 is the final velocity of the projectile after the collision.

The loss of energy in the collision is:

ΔK = K1 - K2

= 187.5 J

This energy is transferred to the blades, causing them to spin. Let's assume that all of this energy is converted to rotational kinetic energy of the turbine.

The final rotational kinetic energy of the turbine is:

K_rot = (1/2) * I * ω^2

where I is the moment of inertia of the turbine and ω is the final rotational velocity of the turbine.

Using the conservation of energy, we can equate the loss of kinetic energy in the collision to the gain in rotational kinetic energy of the turbine:

ΔK = K_rot

187.5 J = (1/2) * 12.5 kg m^2 * ω^2

ω^2 = (2 * 187.5 J) / (12.5 kg m^2)

ω^2 = 30

ω = 5.47 rad/s

Therefore, the rotational velocity of the turbine after the collision is 5.47 rad/s.

User Genfood
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