The slider P can be moved inward by means of the string S , while the slotted ar

Question

The slider P can be moved inward by means of the string S, while the slotted arm rotates about point O. The angular position of the arm is given by ? = 0.55t - 0.032t2, where ? is in radians and t is in seconds. The slider is at r = 1.46 m when t = 0 and thereafter is drawn inward at the constant rate of 0.12 m/s. Determine the velocity v and acceleration a of the slider when t = 4.4 s. Express your answers in the x-y coordinate system.

The slider P can be moved inward by means of the string S, while the slotted arm rotates about point O. The angular position of the arm is given by ? = 0.55t - 0.032t2, where ? is in radians and t is in seconds. The slider is at r = 1.46 m when t = 0 and thereafter is drawn inward at the constant rate of 0.12 m/s. Determine the velocity v and acceleration a of the slider when t = 4.4 s. Express your answers in the x-y coordinate system.

Explanation / Answer

let us denote the angle by A => A= 0.55t-0.032(t^2)

dA/dt = w = 0.55 - 0.064t (angular speed)

dw/dt = alpha = -0.064

at t=4.4s A = 1.80048 rad w= 0.2684 rad/s alpha = -0.064 rad/(s^2)

=> w = 0.2684 k   alpha = -0.064 k

dr/dt = -0.12 m/s (as it is dragged inside)

=> r = -0.12t +c at t=0 r= 1.46m => r= -0.12t + 1.46

=> at t=4.4s r=0.932m

vector, r = r(cosA i + sinA j)

= 0.932(cosA i + sinA j)

= -0.2122 i + 0.9075 j

velocity of P , V = w x r + Vr

where Vr is the realtive velocity of P w.r.t slider

in this case Vr = -0.12(cosA i + sinA j)

= 0.02732 i - 0.11685 j

=> V = (0.2684 k) x ( -0.2122 i + 0.9075 j ) + (0.02732 i - 0.11685 j)

= - 0.05695 j - 0.243573 i + (0.02732 i - 0.11685 j)

= -0.216253 i - 0.1738 j m/s

To find acceleration let us take a moving frame attached to the slider.So the moving frame rotates with same w and alpha of slider.Let P' be a point on the slider that exactly coincides with P at the given instant but does not move relative to the slider.

then ,

acceleration of P, aP = aP' + aP/relative to moving frame+ a,coriolis

aP' is the acceleration of P'

a,coriolis is the coriolis acc. which is 2(wxVr)

aP' = alpha x r - (w^2) r

= -0.064 k x (-0.2122 i + 0.9075 j) - (0.2684^2) (-0.2122 i + 0.9075 j)

= 0.013581 j + 0.05808 i + 0.015287 i - 0.065375 j

= 0.073367 i -0.051794 j

aP/relative = d(Vr)/dt =0 as Vr = -0.12 m/s is constant

a,coriolis = 2(0.2684 k x (0.02732 i - 0.11685 j) )

= 0.014665 j + 0.062725 i

=> aP = 0.073367 i -0.051794 j + 0 + 0.014665 j + 0.062725 i

=( 0.136092 i - 0.037129 j ) m/s^2

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