A straight rack rests on a gear of radius r = 3 in. and is attached to a block B

Question

A straight rack rests on a gear of radius r = 3 in. and is attached to a block B as shown. Knowing that at the instant shown theta = 20 degree, the angular velocity of gear D is 4.2 rad/s clockwise, and it is speeding up at a rate of 2.6 rad/s^2, determine the angular acceleration of AB and the acceleration of block B. The angular acceleration of AB is ___ rad/s^2 clockwise. (Round the final answer to three decimal places.) The acceleration of block B is ___ in/s^2 rightarrow. (Round the final answer to one decimal place.)

Explanation / Answer

let the block B move a distance xin the right direction
then before this motion, tan(theta) = r/D [ where r is radius of the gear, and D is the distance of the center of the gear from the center of block B]
after this motion
tan(theta') = r/(D + x)

diffferentiating wrt time

so, sec^2(theta)d(theta)/dt = -r(dD/dt)/D^2
diffferentiating wrt time again

so, sec^2(theta)d^2(theta)/dt^2 + 2*tan(theta)sec^2(theta)(d(theta)/dt)^2 = r[-D(d^2D/dt^2) + 2*(dD/dt)^2]/D^3
now d(theta)/dt is angular velocity of the rack = w
dD/dt is velocity of the block B = v
d^D/dt^2 is acceleration of the block B = a
d^2(theta)/dt is angular acceleration of the rack = alpha

so the equations become
sec^2(theta)*w = -rv/D^2
sec^2(theta)*alpha + 2*tan(theta)sec^2(theta)w^2 = r[-D*a + 2*v^2]/D^3

also, if angular velocity of the gear D is W
Wr*cos(theta) = v
so, theta = 20 deg
W = 4.2 rad/s
r = 3 in = 0.0762m
so v = 0.3 m/s
but D = r/tan(20) = 0.209 m
so,
sec^2(20)*w = -0.0762*0.3/0.209^2
w = -0.46054 rad/s [ anticlockwise, the angular velocity of the rack]

so, sec^(20)*alpha + 2*tan(20)sec^2(20)*0.46054^2 = 0.0762[-0.209*a + 2*0.3^2]/0.209^3
1.1324*alpha = -1.7444a + 1.32756 .... (1)

also, alpha*r*cos(theta) = a
so, 1.1324*alpha = -1.7444(alpha*0.0762cos(20)) + 1.32756
alpha = 0.9140013 rad/s/s
a = 0.06544 m/s

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