A disk with mass m = 11.5 kg and radius R = 0.3 m begins at rest and accelerates

Question

A disk with mass m = 11.5 kg and radius R = 0.3 m begins at rest and accelerates uniformly for t = 16.7 s, to a final angular speed of = 29 rad/s.

1)

What is the angular acceleration of the disk?
rad/s2

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2)

What is the angular displacement over the 16.7 s?
rad

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3)

What is the moment of inertia of the disk?
kg-m2

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4)

What is the change in rotational energy of the disk?
J

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5)

What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?
m/s2

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6)

What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?
m/s2

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7)

What is the final speed of a point on the disk half-way between the center of the disk and the rim?
m/s

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8)

What is the total distance a point on the rim of the disk travels during the 16.7 seconds?

m

Explanation / Answer

here,

mass , m = 11.5 kg

radius , r = 0.3 m

time taken , t = 16.7 s

final angular velocity , wf = 29 rad/s

1)

let the angular accelration be a

using first equation of motion

wf = 0 + a*t

29 = a*16.7

a = 1.74 rad/s^2

the angular accelration is 1.74 rad/s^2

(2)

the angular displacement , theta = wf^2/(2*a)

theta = 29^2 / ( 2 * 1.74)

theta = 241.67 rad

the angular displacement is 241.67 rad

(3)

the moment of inertia , I = m*r^2/2

I = 11.5 * 0.3^2 /2

I = 0.5175 kg.m^2

the moment of inertia of the disc is 0.5175 kg.m^2

(4)

change in rotation energy of the disk , E = 0.5 * I * wf^2 - 0

E = 0.5 * 0.5175 * 29^2

E = 217.61 J

the change in the rotational energy is 217.61 J

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