Question

A thin rod (length = 1.85 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 the object fixed to the top of the rod. The rod, starting from rest, tips over and rotates downward. (a) What is the angular speed of the rod just before it strikes the floor? (Hint: Consider using the principle of conservation of mechanical energy.)(b) What is the magnitude of the angular acceleration of the rod just before it strikes the floor?

Answer 1

length of the thin rod, L = 1.85 m = h

The mass of the rod may be ignored & compared to the mass of the object fixed to the top of the rod.

initial speed of rod, u = 0 m/s

(a) the angular speed of the rod just before it strikes the floor which is given as :

using conservation of energy,

potential energy = kinetic energy

mgh = 1/2 mv²

v² = 2gh

or v = 2gh                       { eq.1 }

where, g = acceleration due to gravity = 9.8 m/s²

inserting the values in above eq.

v = 2 (9.8 m/s²) (1.85 m)

v = 36.26    m/s

v = 6.02 m/s

using a relation between linear speed & angular speed,

v = r                     { eq.2 }

or = v / r

where, r = radius of the rod = 1.85 m

inserting the values in eq.2

= (6.02 m/s) / (1.85 m)

= 3.25 rad/s

(b) the magnitude of the angular acceleration of the rod just before it strikes the floor which is given as :

using an equation, = _T / r               { eq. 3 }

where, _T = linear tangential acceleration = 9.8 m/s²

At horizontal, the mass 'm' will be accelerating at g = 9.8 m/s²

inserting the values in eq.3,

= (9.8 m/s²) / (1.85 m)

= 5.29 rad/s²