Question

A disk-shaped machine part has a diameter of 39.0 cm. Its angular position is given by

*θ* = −1.22*t*^{3} +
1.60*t*^{2},

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

Answer #1

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

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