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An 80.0 kg merry-go-round of radius 1.20 m is accelerating horizontally from rest by a constant force of 70.0 N applied tangentially to the merry-go-round. Find the kinetic energy of the merry-go-round after 5.00 s. (Assume it is a solid cylinder.)

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onyebuchinnaji

The kinetic energy of the merry-go-round after 5.00 s, when a constant force is applied tangentially to it, is determined as 3,069.5 J.

Conservation of angular momentum

The kinetic energy of the merry-go-round is determined by applying the principle of conservation of angular momentum as follows;

Iα = rF

where;

  • I is moment of inertia of the merry-go-round (solid cylinder)

I = ¹/₂mr²

I = ¹/₂ x 80 x (1.2)²

I = 57.6 kgm²

α = (rF)/I

α = ( 1.2 x 70) / (57.6)

α = 1.46 rad/s²

Tangential acceleration of the merry-go-round

a = αR

a = 1.46 x 1.2

a = 1.752 m/s²

Velocity of the merry-go-round

The velocity of the merry-go-round after 5 s is calculated as follows;

v = u + at

v = 0 + (1.752x 5)

v = 8.76 m/s

Kinetic energy of the merry-go-round

K.E = ¹/₂mv²

K.E = ¹/₂ x 80 x 8.76²

K.E = 3,069.5 J

Learn more about kinetic energy here: https://brainly.com/question/25959744

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