V2 = V,2 + 2aar v +at Rotational kinematics is mathematically very similar to li
ID: 1883448 • Letter: V
Question
V2 = V,2 + 2aar v +at Rotational kinematics is mathematically very similar to linear kinematics. Consider the object shown. Take r = 1.2 m and = 5 rad/s. I. Consider uniform circular motion. In one complete rotation: Distance through which point P on the body moves (angle 8 is in radians) a) How many radians does it traverse? b) How far does it go in meters? c) What is its linear speed? d) What is its period? Linear speed of point P (angular speed o is in rad/o ' by point P II. Consider angular acceleration. Assume an initial angular speed 000-5 radis at t-0 s, with a constant -2 rads a) What is its angular velocity at t = 3.0 s? b) How far has it moved in radians in this time? c) What is its linear velocity at - 3.0 s?? d) What is its tangential acceleration? e) What is its centripetal acceleration at t-3.0s? f) What is the magnitude of its total acceleration? Sketch centripetal, tangential, and total accelerations on the figure above.Explanation / Answer
Solution:
1) a) In one complete revolution 2 pi radians are covered.
1) b) Distance covered in one rotation = 2 * pi * r = 2 * 3.14 * 1.2 = 7.54 m
1) c) Linear speed = v = r * = 1.2 * 5 = 6 m/s
1) d) Time period T = 2*pi / w = 2 * 3.14 * / 5 = 1.256 seconds.
2) a) using the formula
=> w = w0 + alpha * t
=> w = 5 + 2 * 3
=> w = 11 rad/s
2) b) using w^2 - w0^2 = 2 * alpha * s
=> 11^2 - 5^2 = 2 * 2 * s
=> s = 24 radian.
2) c) v = r * w
=> v = 1.2 * 11
=> v = 13.2 m/s
2) d) tangential acceleration = r * alpha = 1.2 * 2 = 2.4 rad/s^2
2) e) centripetal acceleration = w^2 * r = 11^2 * 1.2 = 145.2 rad/s^2
2) f) total acceleration = sqrt( 2.4^2 + 145.2^2 )
=> total acceleration = 145.219 rad/s^2
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