A 4.0 kg block rests at the edge of a disk with a mass of 5.0 kg and a radius of

Question

A 4.0 kg block rests at the edge of a disk with a mass of 5.0 kg and a radius of 30 cm.
(a) A 2 N force is applied to the edge of the disk. What is the angular acceleration of the disk
and block? (Hint: Don't forget to include the moment of inertia of the block. Treat it as a single
particle).
(b) The block is held on the disk by static friction. The maximum force that static friction can
oppose is 24 N. When the block comes off the disk, what is its tangential velocity?
(c) The block flies off the disk how long after the 2 N force is first applied?

Explanation / Answer

a) T = I*

moment of inertia of the system = 4*0.32 + 5 * 0.32/2 = 0.585
T = 2*0.3 = 0.6 Nm
0.6= 0.585*
angular acceleration = = 1.0256 s-2

b) F=ma
24 = 4* a
acceleration = 6 m/s2
now since total acceleration will take into account both tangential ans centrifugal acc.
srqt((R) 2 + (V2/R)2) = 6
0.0946669 + (V2/R)2 = 36
v = tangential velocity = 1.34 m/s

c) 1.34 = 1.0256*0.3 * t
t = 4.357 seconds

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