Please anwer only if you know the answer. and show all work. Ill rate quickly. t

Question

Please anwer only if you know the answer. and show all work. Ill rate quickly. thanks

A uniform ladder with a length l and a mass m is released from rest when it is vertical. If is allowed to fall freely, determine the angle theta at which the bottom (O) begins to lift off the ground. Assume the ladder to be a slender rod. Neglect the friction of the ladder with the ground. (It's an icy day.) Now comes the strange part: Assume the ladder is up against a small step, of negligible height but still tall enough to prevent the ladder from sliding to the left (...if it felt so inclined. You may neglect the free will of the ladder.) This can be solved with Newton's laws alone, but using energy where appropriate cuts down on the algebra. If you really want to have fun1, you might want to determine the angle theta at which the base of the ladder begins to move off to the right. (Of course, you only have my word that any such thing happens.) Now, try to think your way through the whole thing.2

Explanation / Answer

Torque = I*alpha

mg*(L/2)*cos theta = -(mL^2 /3)*alpha

alpha = a / r = a/ (L/2) = 2a / L

mg*(L/2)*cos theta = (mL^2 /3)*(2a / L)

a = (3/4)*g*cos theta

Also, alpha = dw / dt = (dw / d theta)*(d theta / dt) = w*(dw / d theta)

Thus, w dw = -(3/2)(g/L)*cos theta d theta

Integrating, w^2 /2 = -(3/2)*(g/L)*sin theta + C

At theta = 90 deg, we have w = 0.

Thus, C = (3/2)*(g/L)

Thus, w = sqrt [(3g/L) (1 - sin theta)]

Also, N = mg - mrw^2 sin theta + m*(r*alpha)*cos theta

When N = 0 we get g = rw^2 sin theta - (r*alpha)*cos theta

or g = (L/2)*w^2 sin theta - (L/2 *alpha)*cos theta

or g = (L/2)*[(3g/L) (1 - sin theta)] * sin theta + (L/2 *3/2 * (g/L) Cos theta)*cos theta

2/3 = (1 - sin theta)*sin theta + 1/2*cos^2 theta

Solviing this, theta = 19.3 deg

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