As a result of friction, the angular speed of a wheel changes with time accordin

Question

As a result of friction, the angular speed of a wheel changes with time according to d/dt = 0et where 0 and are constants. The angular speed changes from 3.70 rad/s at t = 0 to 2.00 rad/s at t = 8.60 s.

(a) Use this information to determine and 0.

= _______s1

0 = ______rad/s

(b) Determine the magnitude of the angular acceleration at t = 3.00 s.

______rad/s2

(c) Determine the number of revolutions the wheel makes in the first 2.50 s

_______rev

(d) Determine the number of revolutions it makes before coming to rest.

_______rev

Explanation / Answer

given that

w(t)=w0*exp(-sigma*t)

at t=0, w=3.7 rad/sec

==>w0*exp(-sigma*0)=3.7

==>w0=3.7 rad/sec

at t=8.6 seconds, w=2 rad/sec

==>3.7*exp(-sigma*8.6)=2

==>exp(-sigma*8.6)=0.54054
==>-sigma*8.6=ln(0.54054)=-0.615186

==>sigma=0.615186/8.6

==>sigma=0.07153 s^(-1)

part b:

w=w0*exp(-sigma*t)

==>angular acceleration=dw/dt

==>angular acceleration=-sigma*w0*exp(-sigma*t)

==>angular acceleration=-0.07153*3.7*exp(-0.07153*t)

=-0.26466*exp(-0.07153*t) rad/s^2

hence at t=3 seconds, angular acceleration=-0.26466*exp(-0.07153*3)=-0.21355 rad/s^2

c)d(theta)/dt=3.7*exp(-0.07153*t)

==>theta=integration of 3.7*exp(-0.07153*t)*dt

=(3.7/(-0.07153))*exp(-0.07153*t)

=-51.7265*exp(-0.07153*t)

putting limits from t=0 to t=2.5 seconds,

total angle covered=-51.7265*exp(-0.07153*2.5)+51.7625*exp(0)

=8.476 radians

number of revolutions=angle/(2*pi)=1.35

hence the wheel makes 1.35 revolutions in 2.5 seconds

part d:

it will come to rest when w(t)=0

as given exponential function becomes zero at t=infinity, we have to use limit from t=0 to t=infinity

in the expression for theta

hence angle covered before coming to rest

=-51.7625*exp(-0.07153*infinity)+51.7625*exp(-0.07153*0)

=51.7625 radians

number of revolutions=angle /(2*pi)

=8.2382

hence total 8.2382 revolutions are completed.

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