1.) A turntable rotates with a constant 2.25 rad/s^2 angular acceleration. After

Question

1.) A turntable rotates with a constant 2.25 rad/s^2 angular acceleration. After 3.00 s it has rotated through an angle of 30.0 rad. What was the angular velocity of the wheel at the beginning of the 3 second interval?

2.) Find the required angular speed, , of an ultracentrifuge for the radial acceleration of a point 3.00 cm from the axis to equal 5.00×105g (where g is the acceleration due to gravity).

3.) A flat (unbanked) curve on a highway has a radius of 250.0 m . A car rounds the curve at a speed of22.0 m/s . What is the minimum coefficient of static friction that will prevent sliding? Suppose that the highway is icy and the coefficient of friction between the tires and pavement is only one-third what you found in the previous part. What should be the maximum speed of the car so it can round the curve safely?

Explanation / Answer

1)   let angular velocity at beginning of 3 sec interval be Wu. so let q = 30 rad be angle rotated during this 3 sec interval .
since angular acceleration of table j = 2.25 r/s^2 is constant then expression for q can be given by
q = Wu*t + (1/2)*j*t^2 ;similar to linear velocity
; eqn s = U*t + (1/2)*a*t^2
substituting values we get
30 = Wu*3 + (1/2) * 2.25 * 3^2
or Wu = (30 - 10.125)/3
Wu = 6.625 rad/s

2) radial acceleration = ^2R
500000*9.81 = ^2*0.03
^2 = 1.635*10^8
= 12786.71 rad/s
= 2pi*f
f = /(2pi)
f = 2035.07 rps
f = 2035.07*60 rpm = 122104.1 rpm

3)  (a) Friction causes a centripetal acceleration:
F(x) = mv²/r = µmg
µ = v²/gr
= (22.0m/s)² / (9.81m/s²)(250m) = 0.197
v = [µgr]
= [(0.197/3)(9.81m/s²)(250m)]
= 12.7 m/s

Hope this helps.

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