In the figure, point P is on the rim of a wheel of radius 2.0 m

Question: In the figure, point P is on the rim of a wheel of radius 2.0 m. At time t = 0, the wheel is at rest, and P is on the x-axis. The wheel undergoes a uniform counterclockwise angular acceleration of 0.010 rad/s2 about the center O.
(a) At time t = 0, what is the tangential acceleration of P?
(b) What is the linear speed of P when it reaches the y-axis?
(c) What is the magnitude of the net linear acceleration of P when it reaches the y-axis?
(d) How long after starting does it take for P to return to its original position on the x-axis?

Solution:

a) a= alpha*r=0.01*2= 0.02m/s2

b) angular speed when it reaches y= sqrt(2*0.01*pie/2)= 0.177rad/s

so linear speed= 0.177*2= 0.354 m/s

c) linear acceleration= sqrt(tangential^2+ centripetal^2)= sqrt(0.02^2 + 0.177^2/2)= 0.126m/s2

d) to return back at x axis angle= 2 pi so t=sqrt(2*2*pie/alpha(tangential acceleration))= 35.44 sec

where alpha(tangential acceleration)= 0.01 rad/s2

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