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The 100-kg homogeneous cylindrical disk is at rest when the force is applied to a cord wrapped around it, causing the disk to roll. Use the principle of work and energy to determine the angular velocity of the disk when it has turned one revolution.

User Maahd
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Complete question:

The 100-kg homogeneous cylindrical disk is at rest when the force F =500N is applied to a cord wrapped around it, causing the disk to roll. Use the principle of work and energy to determine the angular velocity of the disk when it has turned one revolution (radius of the disk = 300mm).

Answer:

The angular velocity of the disk when it has turned one revolution is 16.712 rad/s

Step-by-step explanation:

From the principle of work and energy;

U = E₂ - E₁, since the disk is initially at rest, T₁ = 0

U = E₂

Work done, U = product of force and perpendicular distance

U = F × d

As the cord winds, force act through the cord at a distance of 2d

U = F × 2d

Distance of one complete revolution = 2πR = 2π(0.3) = 0.6π

U = 500 × 2(0.6π) = 1885.2 J

Kinetic energy E₂
= (1)/(2)I \omega^2 + (1)/(2)m v^2


E_2 = (1)/(2)[((1)/(2)mR^2)\omega^2] + (1)/(2)m(\omega R)^2\\\\E_2 = (1)/(2)[((1)/(2)*100*0.3^2)\omega^2] +(1)/(2)*100(\omega)^2*0.3^2\\\\E_2 = 2.25 \omega^2 +4.5 \omega^2\\\\E_2 = 6.75 \omega^2

Recall that U = E₂

1885.2 = 6.75ω²

ω² = 1885.2/6.75

ω² = 279.2889

ω = √279.2889

ω = 16.712 rad/s

Therefore, the angular velocity of the disk when it has turned one revolution is 16.712 rad/s

User Faylon
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