The balance wheel of a watch oscillates with angular amplitude 1.0π rad and period 0.33 s. Find (a) the maximum angular speed of the wheel, (b) the angular speed of the wheel at displacement 1.0π/2 rad, and (c) the magnitude of the angular acceleration at displacement 1.0π/4 rad.
ans)
from above data that
we know that
x = A*cos wt
1.
V max = A*w
w = 2*pi/T
V max = 2*pi*A/T
V max = 2*pi*1.0*pi/0.33 =59.75 rad/sec
B.
now calculate the
x = A*cos wt
at given x
1.0*pi/2 = 1.0*pi*cos wt
cos wt = 1/2
wt = pi/3
So,
V = A*w*sin wt
V = 59.75*sin 60 deg
V = 51.745 rad/sec
C.
at given x,
1.0*pi/4 = 1.0*pi*cos wt
cos wt = 1/4,
Now acceleration is given by:
a = -A*w^2*cos wt
a = -1.0*pi*(2*pi/0.33)^2*(1/4)
a = -284.29 rad/sec^2
magnitude of angular acceleration is asked, So a = +284.29 rad/sec^2
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