Consider a playground merry-go-round with mass 249 kg and radius 1.52 m. Approxi

Question

Consider a playground merry-go-round with mass 249 kg and radius 1.52 m. Approximate the merry-go-round as a solid, flat cylinder rotating about its axis of symmetry; you can look up the moment of inertia for this geometry in any college physics textbook. (E.g., the OpenStax College Physics figure captioned"Some rotational inertias".)

(a) Calculate the rotational kinetic energy in the merry-go-round when it has an angular velocity of 22.0 rpm.

b) Find the number of revolutions you would have to push the merry-go-round at the edge with force 21.5 N to achieve this angular velocity starting from rest.

(c) Calculate the force you would need to exert to stop the merry-go-round in 4 revolutions.

In 2010, a large rotary saw blade came loose from a street saw, tore across a lawn, and cut into a house. (See YouTube video [+].)

Suppose the saw blade has a diameter of 61 cm, a mass of 8.4 kg, and rotates at 3.8 revolutions/sec as it rolls across the lawn.

What is the saw blade's angular momentum?

When the saw blade encounters the house, a net force applied tangent to the edge of the blade slows its rotation. (A radial force stops the linear motion of its center of mass, but here we are concerned with the tangent force that slows its rotation.) Suppose the net force tangent to the edge of the blade is 22.2 N. How long does it take the saw blade to come to a stop?

Explanation / Answer

here,

mass , m = 249 kg

radius , r = 1.52 m

(a)

angular velocity , w = 22 rpm

w = 2.3 rad/s

the rotation kinetic energy , KE = 0.5 * I * w^2

KE = 0.5 * 0.5 * 249 * 1.52^2 * 2.3^2

KE = 760.82 J

the rotational kinetic energy of the merry go ground is 760.82 J

(b)

F = 21.5 N

let the angular accelration be alpha

F*r = I*alpha

21.5 * 1.52 = 0.5 * 249 * 1.52^2 * alpha

alpha = 0.114 rad/s^2

let theta be the angle

using third equation of motion

2*alpha*theta = 2.3^2

theta = 23.28 rad

the number of revolutions are 3.7 revolutions

(c)

when theta2 = 4 * 6.28 rad

theta2 = 25.12 rad

let the accelration be alpha'

using third equation of motion

2*alpha' * 25.12 = 2.3^2

alpha' = 0.105 rad/s^2

the force exerted , F' = I*alpha'/r

F' = 0.5 * 249 * 1.52^2 * 0.105 /1.52

F' = 19.35 N

the force exerted to stop the merry go ground is 19.35 N

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