A disk-shaped machine part has a diameter of 39.0 cm. Its angular position is given by
θ = −1.22t3 + 1.60t2,
where t is in seconds and θ is in radians.
(a) What is the maximum angular speed of the part during this
time interval? (Assume the time interval is from t = 0 to
when the part reverses its direction.)
rad/s
(b) What is the maximum tangential speed of a point halfway to the
rim of the part during this time interval?
m/s
(c) What is the time t for which the part reverses its
direction of motion after
t = 0?
s
(d) Through how many revolutions has the part turned when it
reaches the time found in part (c)?
revolutions
a) given
then angular speed is
and angular acceleration is
a) Angular speed is maximum when a = 0
Therefore
or
b) diameter is d = 0.39 m
the maximum tangential speed of a point d/4 at t = 0.437 s is
c) the time for which the part reverses its direction of motion is when theta = 0
or
d) The total angular displacement during this time is
The number of revolutions is
Get Answers For Free
Most questions answered within 1 hours.