A uniform disk with mass 44.1 kg and radius 0.230 m is pivoted at its center abo

Question

A uniform disk with mass 44.1 kg and radius 0.230 m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 34.0 N is applied tangent to the rim of the disk.

A. What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.150 revolution?

B. What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.150 revolution?

Explanation / Answer

Here ,

m1 = 44.1 Kg

r = 0.230 m

Force, F = 34 N

A)

let the moment of inertia is I

I = 0.5 * m* r^2

I = 0.5 ( 44.1) * 0.230^2

I = 1.17 Kg.m^2

let the angular acceleration is a

Using second law of motion

I * a = F * r

1.17 * a = 34 * 0.230

a = 6.704 rad/s^2

let the angular velocity is wi

Using third equation of motion

wi^2 = 2 * theta * a

wi^2 = 2 * 0.15 * 2pi * 6.704

wi = 3.55 rad/s
tangential velocity = wi * r

tangential velocity = 3.55 * 0.23 = 0.82 m/s

the tangential velocity is 0.82 m/s

b)

centripetal acceleration , ac = v^2/r

ac = 0.82^2/.23

ac = 2.91 m/s^2

tangential accelertion , at = a * r

at = 6.704 * 0.23

at = 1.541 m/s^2

resultant acceleration = sqrt(at^2 + ac^2)

resultant acceleration = sqrt(1.541^2 + 2.91^2)

resultant acceleration = 3.29 m/s^2

the resultant acceleration is 3.29 m/s^2

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