- #1
nbroyle1
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A 36.5-cm diameter disk rotates with a constant angular acceleration of 2.00 rad/s2. It starts from rest at t = 0, and a line drawn from the center of the disk to a point P on the rim of the disk makes an angle of 57.3° with the positive x-axis at this time.
(a) Find the angular speed of the wheel at t = 2.30 s.
(b) Find the linear velocity and tangential acceleration of P at t = 2.30 s.
(c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.
(a) to solve for angular speed W=at or W=(2)(2.3) and got 4.6m/s
(b) first I calculated the radius by multiplying 36.5*.01 then dividing by 2 and got .1825.
next I plugged it into V=(.1825)(4.6) to solve for linear velocity and got .8395.
To get tangential acceleration I used A=(.1825)(2) and got .365.
(c) θ=1+.5(2)(2.3)^2 and got 6.29 radians which I converted to degrees and got 360.39 and subtracted 360 to get position with respect to positive x-axis.
Not sure If I did part this right can I get some verification??
(a) Find the angular speed of the wheel at t = 2.30 s.
(b) Find the linear velocity and tangential acceleration of P at t = 2.30 s.
(c) Find the position of P (in degrees, with respect to the positive x-axis) at t = 2.30s.
(a) to solve for angular speed W=at or W=(2)(2.3) and got 4.6m/s
(b) first I calculated the radius by multiplying 36.5*.01 then dividing by 2 and got .1825.
next I plugged it into V=(.1825)(4.6) to solve for linear velocity and got .8395.
To get tangential acceleration I used A=(.1825)(2) and got .365.
(c) θ=1+.5(2)(2.3)^2 and got 6.29 radians which I converted to degrees and got 360.39 and subtracted 360 to get position with respect to positive x-axis.
Not sure If I did part this right can I get some verification??