A wheel starts from rest and rotates with a constant angular acceleration about

Question

A wheel starts from rest and rotates with a constant angular acceleration about a fixed axis. It completes the first revolution 6.0 s after it started. How long after it started will the wheel complete the second revolution? A rectangular plate is rotating with a constant angular acceleration about an axis that passes perpendicularly through one comer, as the drawing shows. The tangential acceleration measured at corner A has twice the magnitude of that measured at corner B. What is the ratio L_1/L_2 of the lengths of the sides of the rectangle? A uniform disk of radius 1.2 m and mass 0.60 kg is rotating at 25 rad/s around an axis that passes through its center and is perpendicular to the disk. A rod makes contact with the rotating disk with a force of 4.5 N at a point 0.75 m from the axis of rotation as shown. The disk is brought to a stop in 5.0 s. What is the coefficient of kinetic friction for the two materials in contact?

Explanation / Answer

1. using second equation of motion in angular terms we have from given condition that:

2 = 0 + 1/2**62   where is the constant angular acceleration.

so = /9 rad/s2   now suppose to complete two revolutions ( rotation by 4) it takes total of time 't'

then 4 = 0 + 1/2**t2   = 1/2*/9*t2   or t = 8.48 s

hence the time taken to complete the second revolution = 8.48 - 6 = 2.48 s

2.    tangential acceleration of point A (aA) = OA* where O is the point about which the plate is rotating and is the constant angular acceleration

then tangential acceleration of point B (aB) = OB*

we are given aA/aB  = 2 or OA = 2*OB or (L12 + L22)0.5 = 2*L1   or L22 = 3*L12

hence L1/L2 = 1/30.5 = 0.574

3. Given: r = 1.2 m; m = 0.6 Kg ; initial angular velocity = 25 rad/s ;

Now when F= 4.5 N was applied at a distance of 0.75 m from the center of rotation then frictional force acting on the disc f = *4.5 where is the coefficient of kinetic friction

also the torque acting on the disc = -f*0.75 = -3.375* (direction of torque is opposite to the direction of motion)

Now applying third equation of motion in angular terms we have   

0 = + t or angular acceleration = -/t = -25/5 = -5 rad/s2

also = I* where the moment of inertia I = 1/2*m*r2 (for disc)

therefore -3.375* = 1/2*0.6*1.22*(-5) or = 0.64

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