A disk with mass m = 8.5 kg and radius R = 0.35 m begins at rest and accelerates

Question

A disk with mass m = 8.5 kg and radius R = 0.35 m begins at rest and accelerates uniformly for t = 18.9 s, to a final angular speed of ? = 29 rad/s.

a) What is the angular acceleration of the disk?

b) What is the angular displacement over the 18.9 s?

c) What is the moment of inertia of the disk?

d) What is the change in rotational energy of the disk?

e) What is the tangential component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?

f) What is the magnitude of the radial component of the acceleration of a point on the rim of the disk when the disk has accelerated to half its final angular speed?

g) What is the final speed of a point on the disk half-way between the center of the disk and the rim?

h) What is the total distance a point on the rim of the disk travels during the 18.9 seconds?

Explanation / Answer

given

m = 8.5 kg, R = 0.35 m

at t1 = 0, w1 = 0

at t2 = 18.9 s, w2 = 29 rad/s

a) angular acceleration, alfa = (w2 - w1)/t

= (29 - 0)/18.9

= 1.53 rad/s^2

b) theta = wo*t + (1/2)*alfa*t^2

= 0 + (1/2)*1.53*18.9^2

= 273 radians

c) I = 0.5*m*R^2

= 0.5*8.5*0.35^2

= 0.521 kg.m^2

d) delta_KE = (1/2)*I*(w2^2 - w1^2)

= (1/2)*0.521*(29^2 - 0^2)

= 219 J

e) Torque = I*alfa

F_tan*r = I*alfa

F_tan = I*alfa/r

= 0.521*1.53/0.35

= 2.28 N

f) a_rad = r*w^2

= 0.35*(29/2)^2

= 73.6 m/s^2

g) v = (r/2)*wf

= (0.35/2)*29

= 5.08 m/s

h) use, s = r*theta

= 0.35*273

= 95.6 m

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