Question

A wheel 1.50 m in diameter lies in a vertical plane and rotates
about its central axis with a constant angular acceleration of 3.95
rad/s^{2}. 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

rad/s

(b) the tangential speed of the point *P*

m/s

(c) the total acceleration of the point *P*

magnitude | m/s^{2} |

direction | ° with respect to the radius to point P |

(d) the angular position of the point *P*

rad

Answer #1

Given,

Initial angular velocity , wi=0

Time , t= 2 s

Angular acceleration, @=3.95 rad/s²

Radius, r=d/2=1.5/2=0.75 m

Initial position. #i=57.3°(π÷180°)=1 rad

(A) the object under the constant angular acceleration, the final angular speed,

wf= wi +@t

=(0)+(3.95)(2)

wf=7.9 rad/s

(B) The tangential speed,

v=r *wf

v=5.925 m/s

(C) The tangential acceleration at P,

a(t)=r* @

=0.75*3.95

=2.9625 m/s²

The radial acceleration at P,

a(r)=v²/r

=(5.925)²/0.75

a(r)=46.8075 m/s²

The Total acceleration of the wheel at P,

a=√{a(t)²+a(r)²}

a=√{(2.9625)²+(46.8075)²}

a=√{2199.72}

a=46.9 m/s

The direction of the acceleration with the radial acceleration vector,

Angle=tan–¹{a(t)/a(r)}

=tan–1{2.9625/46.8075}

Angle=3.62°

(D) The final angular position at P,

#f=#i+wi+½@t²

#f =1rad +0+½*(3.95)(2)²

=1 rad +7.9 rad

=8.9 rad

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