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2 years ago

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A 235-kg merry-go-round at the Great Escape in Lake George is in the shape of a uniform, solid, horizontal disk of radius 1.50 m

. It is set in motion by wrapping a rope about the rim of the disk and pulling on the rope. What constant force would have to be exerted on the rope to bring the merry-go-round from rest to an angular speed of 0.400 rev/s in 2.00 s?

1 answer:

Kruka [31]2 years ago

4 0

Answer:

The constant force to be exerted on the rope is 221.55 N

Explanation:

Given;

mass of the merry, m = 235 kg

radius, r = 1.5 m

number of revolution per second, = 0.4 rev/s

time of motion, t= 2.00 s

The angular acceleration is given by;

$\alpha = \frac{0.4 \ rev}{s} *\frac{2\pi \ rad}{rev} *\frac{1}{2.0\ s} = 1.257 \ rad/s^2$

Torque is given by;

τ = F.r

Also torque in uniform solid disk is given by;

τ = ¹/₂mr²α

Thus, equating the two;

F.r = ¹/₂mr²α

F = ¹/₂mrα

F = ¹/₂(235)(1.5)(1.257)

F = 221.55 N

Therefore, the constant force to be exerted on the rope is 221.55 N

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