A wheel 1.65 m in diameter lies in a vertical plane and rotates about its central axis with a constant angular acceleration of 4.35 rad/s2. The wheel starts at rest at t = 0, and the radius vector of a certain point P on the rim makes an angle of 57.3° with the horizontal at this time. At t = 2.00 s, find the following.
(a) the angular speed of the wheel
(b) the tangential speed of the point P
(c) the total acceleration of the point P
magnitude | |
direction |
(d) the angular position of the point P
a )
given
= 4.35 rad/s2
t = 2 sec
the wheel is started at rest so o = 0
= o + t
= t
= 4.35 X 2
= 8.7 rad/sec
b )
given
d = 1.65 m
r = d / 2 = 0.825 m
we have v = r
v = 0.825 X 8.7
v = 7.177 m/sec
c )
the tangential acceleration
at = dv / dt
= d ( r ) / dt
at = r
at = 0.825 X 4.35 = 3.588 m/sec2
ar = v2 / r
ar = 7.1772 / 0.825
ar = 62.43 m/sec2
a = ( at2 + ar2 )1/2
a = 62.533 m/sec2
= tan-1 ( 3.588 / 62.43)
= 3.280
d )
= o + o t + 1/2 t2
= 57.3 X 3.14 / 180 + 0 + 0.5 X 4.35 X 4
= 0.99 + 8.7
= 9.69 rad
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