Query 31

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course PHY 231

11/21 1:07 pm

031. `query 31*********************************************

Question: `qUniv. 10.52 (10.44 10th edition). 55 kg wheel .52 m diam ax pressed into wheel 160 N normal force mu =.60. 6.5 m N friction torque; crank handle .5 m long; bring to 120 rev/min in 9 sec; torque required? Force to maintain 120 rev/min? How long to coast to rest if ax removed?

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Your solution:

r = 0.26 m

120 rpm= 4π rad/s -----> alpha = 4π/9 rad/s^2

I = 1/2MR^2 = 1/2(50)(0.26)^2 = 1.69 kg m^2

tauNet = 1.69*4π/9 = 2.36 mN

Nf= 160 N-----> ƒ= 0.6(160)= 96 N------>tauƒ= (96*0.26)+6.5 = 31.46 mN

(a)tauNet = 2.36 = tauCrank-31.46-----> tauCrank = 33.82 = Fhand*0.5 -----> Fhand= 67.64 N

(b) tauNet = 0 = tauCrank -31.46----->Fhand = 62.92 N

Confidence rating:3

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Given Solution:

`a** The system is brought from rest to a final angular velocity of 120 rev/min * 1min/60 sec * 2`pi/1 rev = 12.6 rad/s.

The angular acceleration is therefore

alpha = change in omega / change in t = 12.6 rad/s / (9 sec) = 1.4 rad/s^2, approx..

The wheel has moment of inertia I = .5 m r^2 = .5 * 55 kg * (.52 m)^2 = 7.5 kg m^2, approx..

To achieve the necessary angular acceleration we have

tauNet = I * alpha = 7.5 kg m^2 * 1.4 rad/s^2 = 10.5 m N.

The frictional force between ax and wheel is .60 * 160 N = 96 N at the rim of the wheel, resulting in torque

tauFrictAx = -96 N * .52 m = -50 m N.

The frictional torque of the wheel is in the direction opposite motion and is therefore

tauFrict = -6.5 m N.

The net torque is the sum of the torques exerted by the crank and friction:

tauNet = tauFrictAx + tauFrict + tauCrank so that the torque necessary from the crank is

tauCrank = tauNet - tauFrict - tauCrank = 10.5 m N - (-50 m N) - (-6.5 m N) = 67 m N.

The crank is .5 m long; the force necessary to achive the 60.5 m N torque is therefore

F = tau / x = 67 m N / (.5 m) = 134 N.

If the ax is removed then the net torque is just the frictional torque -6.5 m N so angular acceleration is

alpha = -6.5 m N / (7.5 kg m^2) = -.84 rad/s^2 approx.

Starting at 120 rpm = 12.6 rad/s the time to come to rest will be

`dt = `dOmega / alpha = -12.6 rad/s / (-.84 rad/s^2) = 14.5 sec, approx.. **

Self-critique (if necessary): The process of the given solution is right but 0.52 is the diameter not the radius.

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