I ONLY NEED THE SECOND PART OF B ANSWERED. ALL OTHERS ARE CORRECT. THANK YOU! A
ID: 1869074 • Letter: I
Question
I ONLY NEED THE SECOND PART OF B ANSWERED. ALL OTHERS ARE CORRECT. THANK YOU!
A wheel 1.70 m in diameter lies in a vertical plane and rotates with a constant angular acceleration of 4.10 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
=8.2 rad/s
(b) the tangential speed and the total acceleration of the point P
=6.97 m/s
=_______ m/s^2
(c) the angular position of the point P
=9.2 rad
Explanation / Answer
a)
w = w0 + at
w = 0 + 4.1*2 = 8.2 rad/s
b)
tangential speed = v = r*w = (1.7/2)*8.2 = 6.97 m/s
tangential acceleration = at = r*alpha = (1.7/2)*4.1 = 3.485 m/s^2
radial acceleration = ar = v^2/r = (6.97)^2/(1.7/2) = 57.154 m/s^2
total acceleration = sqrt(at^2 + ar^2) = sqrt(3.485^2 + 57.154^2) = 57.26 m/s^2
d)
theta = theta0 + w0*t + 0.5*a*t^2
theta = 57.3*(pi/180) + 0 + 0.5*4.1*2^2 = 9.2 rad
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