Now consider the same bar as in Ex. 3 except with the vertical wall removed Agai

Question

Now consider the same bar as in Ex. 3 except with the vertical wall removed Again suppose that the bar is released from rest with = 30°. Note: You may a) Express angular acceleration 1.0. Of g and 1 at the moment just after b) Show that the vertical force that the ground exerts on the bar just after it c) Derive an expression for the angular acceleration = a(t) after the bar is find it necessary to use the same set of equations to solve parts (a) and (b) release. [10 points] is released is Fv-4mg/7. [5 points] released from rest. This should be in the form of a 2nd-order ODE only in terms of m, g, 1 and . [10 points] d) Now suppose that the roller also has a mass m. What is its velocity when the bar reaches horizontal (ie. = 90°)? [5 points]

Explanation / Answer

given

theta = 30 deg

let mass of rod be m

a. let initial angular acceleration be alpha

then, mg*(l/2)sin(theta) = ml^2*alpha/3

alpha = 3g*sin(theta)/2l = 0.85714g/l

b. let the vertical force by the ground be F

then

mg - F = m*a

but a = alpha*l/2 = 0.75g/2 = 0.375g

hence

F = m(g - 0.42857g) = 4mg/7

c. at any time t

mg*(l/2)sin(theta) = ml^2*alpha(t)/3

alpha(t) = 3gsin(theta)/2l

d. mass of roller = m

when the bar becomes horizontal

from conservation of energy

0.5(2m)v^2 = mglcos(theta)/2

mv^2 = glcos(30)/2

v = sqroot(lcos(30)*g/2)

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