dynamics problem Pin Pis attached to the wheel shown and slides in a slot cut in

Question

dynamics problem

Pin Pis attached to the wheel shown and slides in a slot cut in bar BD. The wheel rolls to the right without slipping with a constant angular velocity of 20.8 rad/s. Knowing that x= 480 mm when O= 0, determine the angular velocity of the bar and the relative velocity of pin P with respect to the rod for the given data. A40 mm 200 mm References eBook & Resources Section Break Difficulty: Medium value: 10.00 points Required information Given: e = 0. (Round the final answers to two decimal places.) The angular velocity of the bar is rad/s U. The relative velocity of pin P with respect to the rod is m/s &

Explanation / Answer

for the given diagram
w = 20 rad/s ( wheel's angular speed)
x = 480 mm = 0.48 m when theta = 0 deg

let angular velocity of the bar be d(phi)/dt
anf Vp relative to the bar be vp

then
vp = dl/dt
l^2 = (x + r*(sin(theta)))^2 + (r*cos(theta))^2
l^2 = x^2 + r^2*sin^2(theta) + 2xrsin(theta) + r^2*cos^2(theta)
l^2 = x^2 + r^2 + 2xr*sin(theta)
2l*dl/dt = 2x*dx/dt + 2r*dr/dt + 2xrcos(theta)*d(theta)/dt + 2sin(theta)r*dx/dt
dx/dt = 20.8*r
2*sqrot(x^2 + 0.14^2)dl/dt = 2x*20.8*r + 2*0.48*r*cos(0)20 + 2*sin(0)*r*dx/dt
2*sqroot(0.48^2 + 0.14^2)dl/dt = 2*0.48*20.8*0.2 + 2*0.48*0.2*20
dl/dt = 7.8336 m/s

also
lcos(phi) = x
dl/dtcos(phi) - sin(phi)l*d(phi)/dt = dx/dt
7.8336*cos(phi) - sin(phi)*sqroot(0.48^2 + 0.14^2)*w' = 20.8*0.2

also
cos(phi) = 0.48/sqroot(0.48^2 + 0.14^2) = 0.96
phi = 16.26 deg
w' = 24 rad/s

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