A uniform disk with mass 40.9 kg and radius 0.300 m is pivoted at its center abo

Question

A uniform disk with mass 40.9 kg and radius 0.300 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 35.0 N is applied tangent to the rim of the disk.

- What is the magnitude v of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.100 revolution?

- What is the magnitude a of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.100 revolution?

Appreciate the help!

Explanation / Answer

given that

F = 35.0 N

r = 0.300 m

m = 40.9 kg

theta = 2*pi*0.100 revolution = 0.628 radians

let

torque = t

tangential velocity = v

angular speed = W

we know that

Torque t = F*r

t = 35*0.3 = 10.5 N*m

we also know that

t = I * alpha

where alpha is the angular acceleration

I = moment of inertia

I for disc = m*r^2 / 2

I = 40.9*(0.3)^2 / 2 = 1.84 kg*m^2

so alpha = t / I

alpha = 10.5 / 1.84= 5.70 rad/s^2

part (a)

Wf = sqrt (2 * alpha * theta)

Tangential velocity v = r * Wf

v = r * sqrt (2* alpha *theta)

v = 0.3 *sqrt (2 * 5.70 * 0.628)

v = 0.80 m/s

part (b)

tangential acceleration aT = r* alpha

aT = 0.3 * 5.70 = 1.71 m/s^2

radial acceleration aR = v^2 / r

aR = (0.80)^2 / .3 = 2.13 m/s^2

resultant acceleration a = sqrt( (aT ^2) + (aR^2) )

a = sqrt ( (1.71)^2) + (2.13)^2) )

a = sqrt (2.92 + 4.54) = 2.73 m/s^2

so resultant acceleration a = 2.73 m/s^2

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