Please show all work and equations used, and step by step for each answer. An 8-

Question

Please show all work and equations used, and step by step for each answer.

An 8-kg thin disk of radius of 0 2 m is initially at rest. A 100 N force is applied parallel to the edge for 8 seconds. (Torque = +20 N-m) Frictional resistance to rotation (frictional torque) is -15 N-m and exists while the disk is How many revolutions will the disk make in 8 seconds? How many total revolutions will the disk make until it stops? For a thin disk. I = 1/2 mr^2 = 1/2 (8kg) (0.2m)^2 = 0.160 kg-m^2 tau_ net = I alpha. alpha must be in (rad)/s^2 omega = omega_0 + alpha t theta = omega_0 t + 1/2 alpha t^2

Explanation / Answer

Torque due to 100 N is T1 = r*F*sin(90) = 0.2*100 = 20 N-m

Frictional torque is T2 = 15 N-m

Net torqu is Tnet = T1-T2 = 20-15 = 5 N-m

Tnet = I*alpha

I = 0.16 kg-m^2

then 5 = 0.16*alpha

alpha = 31.25 rad/s^2

Using theta = (wo*t)+(0.5*alpha*t^2)

theta = (0*t)+(0.5*31.25*8^2) = 1000 rad

revolutions are n = 1000/(2*3.142) = 159 revolutions

time taken before stopping is t

from 0sec to 8 sec

then using w = wo+(alpha*t)

W = 0 +(31.25*8) = 250 rad/sec

Now using W = Wo+(alpha*t)
after 8 sec ,there is no 100 N applied force ,hence

T2 = I*alpha

alpha = 15/0.16 = 93.75 rad/s^2

0 = 250 -(93.75*t)

t = 2.67 sec

So now using

theta = (wo*t)+(0.5*alpha*t^2)

theta = (250*2.67) - (0.5*93.75*2.67^2)

theta = 333.34 rad

no.of revolutions are 333.34/(2*3.142) = 53 rev

total revoltions are 159+53 = 212 revolutions

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