A uniform disk with mass 40.0kg and radius 0.260m is pivoted at its center about

Question

A uniform disk with mass 40.0kg and radius 0.260m is pivoted at its center about a horizontal, frictionless axle that is stationary. The disk is initially at rest, and then a constant force 33.0N is applied tangent to the rim of the disk.

Part 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 0.290 revolution?

Part 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 0.290 revolution?

Explanation / 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*33N/(40 kg * 0.26m) = 6.34 rad/s/s

0.29 rev = 1.8 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 = 1.8 rad

wf^2=0+2*6.34*1.8 => wf = 4.77 rad/s

this is the angular velocity, the linear velocity = w r =1.24 m/s

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

the linear acceleration = alpha r = 6.34rad/s/s x 0.26m = 1.64 m/s/s

a centrip = v^2/r = 1.24^2/0.26 = 5.92m/s/s

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

resultant accel =6.13m/s2

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