A thin rod (length = 1.50 m) is oriented vertically, with its bottom end attache

Question

A thin rod (length = 1.50 m) is oriented vertically, with its bottom end attached to the floor by means of a frictionless hinge.The mass of the rod may be ignored, compared to the mass of theobject fixed to the top of the rod. The rod, starting from rest,tips over and rotates downward. (a) What is theangular speed of the rod just before it strikes the floor?(Hint: Consider using the principle of conservation ofmechanical energy.) (b) What is the magnitudeof the angular acceleration of the rod just before it strikes the floor?

I got the answer for (a). It's 3.61 rad/s. (b) is 6.53 rad/s², but I don't know how to get it. Please help. I have a test in a couple of days.

Explanation / Answer

a) PE=KE
     mgh = mv*v/2
m falls away so
         gh = v*v/2
    and g = 9.81
so
   9.81*1.5*2 = v*v
                v = 5.43m/s
then ang speed = v/r = 5.43/1.5 = 3.61rad/s

b) and to find ang acc:

90degrees = pi/2 and s = r*pi/2 = 1.5*pi/2 = 2.36m
so
9.81m/s * ( (pi/2) rad / 2.36m ) = 6.53 rad/s

Related Questions

Navigate

• Browse (All)
• Browse A
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.