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andrew11 [14]
3 years ago
12

A merry-go-round on a playground consists of a horizontal solid disk with a weight of 805 N and a radius of 1.47 m. A child appl

ies a force 49.0 N tangentially to the edge of the disk to start it from rest. What is the kinetic energy of the merry-go-round disk (in J) after 2.95 s?
Physics
1 answer:
snow_tiger [21]3 years ago
8 0

Answer:

255.34 J

Explanation:

Given,

Weight of disk = 805 N

radius = 1.47 m

Force applied by the child = 49 N

time = 2.95 s

KE = ?

mass of the disk

M = \dfrac{W}{g}= \dfrac{805}{9.81} = 82.059\ Kg

Moment of inertia of the disk

I = \dfrac{1}{2}Mr^2

I = \dfrac{1}{2}\times 82.059\times 1.47^2 =88.66\ kgm^2

Torque on the child

\tau = F \times r = 49 \times 1.47 = 72.03 Nm

Angular acceleration

\alpha = \dfrac{\tau}{I}=\dfrac{72.03}{88.66} = 0.812\ rad/s^2

So, angular speed at t = 2.95 s

\omega = \alpha t = 0.812 \times 2.95 = 2.4\ rad/s

Now, KE of the merry go round

KE = \dfrac{1}{2} I \omega^2 = \dfrac{1}{2}\times 88.66\times 2.4^2 = 255.34 J

Hence, the Kinetic energy of the merry go round = 255.34 J

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