A uniform disk with mass 43.5 and radius 0.250 is pivoted at its center about a

Question

A uniform disk with mass 43.5 and radius 0.250 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 is applied tangent to the rim of the disk.

What is the magnitude of the tangential velocity of a point on the rim of the disk after the disk has turned through 0.130 revolution?

What is the magnitude of the resultant acceleration of a point on the rim of the disk after the disk has turned through 0.130 revolution?

Explanation / Answer

solved a similar question already with different numbers.. i put the question and the answer.. hope it helps!! :)

A uniform disk with mass 35.2 kg and radius 0.200 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 28.5 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 0.340 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 0.340 revolution?

ANSWER:

for the first part, use torque = I alpha where I is the moment of inertia and alpha is the angular acceleration

the moment of inertia of a soid disk is 1/2 MR^2

the torque is F x R since the force is applied tangentially to the rim of the disk, so we have

F R = 1/2 MR^2 alpha

solve for alpha: alpha = 2 F/M R = 2*28.5N/(35.2 kg * 0.2m) = 8.10 rad/s/s

0.34 rev = 2.14 rad

find angular velocity from

wf^2 = w0^2 + 2 alpha theta

wf, w0 = final, initial ang vel
alpha = ang accel = 8.10rad/s/s
theta = angular displacement = 2.14 rad

wf^2=0+2*8.10*2.14 => wf = 5.8 rad/s

this is the angular velocity, the linear velocity = w r = 1.18m/s

resultant accel = sqrt[a centip^2 + a tangential^2]

the linear acceleration = alpha r = 8.1rad/s/s x 0.2m = 1.6 m/s/s

a centrip = v^2/r = 1.18^2/0.2 = 6.92m/s/s

substitute these into the resultant accel equation and solve for the resultant accel in m/s/s

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