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3 years ago

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A 416 kg merry-go-round in the shape of a horizontal disk with a radius of 1.7 m is set in motion by wrapping a rope about the r

im of the disk and pulling on the rope. How large a torque would have to be exerted to bring the merry-go-round from rest to an angular speed of 3.7 rad/s in 2.9 s

1 answer:

kipiarov [429]3 years ago

6 0

Answer:

The torque exerted on the merry-go-round is 766.95 Nm

Explanation:

Given;

mass of the merry-go-round, m = 416 kg

radius of the disk, r = 1.7 m

angular speed of the merry-go-round, ω = 3.7 rad/s

time of motion, t = 2.9 s

The torque exerted on the merry-go-round is calculated as;

$\tau = Fr= I\alpha\\\\\tau = (\frac{1}{2} m r^2)(\frac{\omega }{t} )\\\\\tau = (\frac{1}{2} \times 416 \times 1.7^2)( \frac{3.7}{2.9} )\\\\\tau = 766.95 \ Nm$

Therefore, the torque exerted on the merry-go-round is 766.95 Nm

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