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hichkok12 [17]
3 years ago
10

A horizontal 649 N merry-go-round of radius 1.05 m is started from rest by a constant horizontal force of 61.3 N applied tangent

ially to the merry-go-round. Find the kinetic energy of the merry-goround after 3.34 s. The acceleration of gravity is 9.8 m/s 2 . Assume the merry-go-round is a solid cylinder. Answer in units of J.
Physics
1 answer:
nexus9112 [7]3 years ago
4 0

Answer:

The kinetic energy of the merry-go-round is 632.82 J

Explanation:

Given;

weight of the merry-go-round, W = 649 N

radius of the merry-go-round, r = 1.05 m

applied horizontal force, F =  61.3 N

acceleration due to gravity, g = 9.8 m/s²

mass of  merry-go-round, m = W/g

                                               = 649/9.8  = 66.225 kg

moment of inertia of merry-go-round, I = ¹/₂mr²

                                                                 = ¹/₂ x 66.225 x (1.05)²

                                                                 = 36.507 kg.m²

Angular acceleration of the merry-go-round, α

τ = Iα = Fr

α = Fr / I

Where;

α is angular acceleration

α = (61.3 x 1.05) / 36.507

α = 1.763 rad/s²

Angular velocity of the merry-go-round, ω

ω = αt

ω = 1.763 x 3.34

ω = 5.888 rad/s

Finally, the kinetic energy of the merry-go-round, K.E

K.E = ¹/₂Iω²

K.E = ¹/₂ x 36.507 x (5.888)²

K.E = 632.82 J

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the magnitude of the shortest displacement from the starting point A to point the final position C is given by:

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As the vector AC is coordinates lie in the third quadrant of the cartesian vector plane whose angle with the west will be positive in the north direction.

The angle of the shortest line connecting the starting point and the final position measured north of west is given by:

\theta = \tan^{-1}(\dfrac{11.85}{41.27})\\\Rightarrow \theta = 16.02^\circ

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