Question

A wheel 1.65 m in diameter lies in a vertical plane and rotates about its central...

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

Homework Answers

Answer #1

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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