Question

Problem 3. A table consists of a horizontal thin uniform circular disk of radius R and mass M, supported on its rim by three thin vertical legs that equidistant from each other. The are acceleration due to gravity is g. suddenly removed. Find the force exerted by the third leg on (a) Suppose two of the legs are the table top immediately after. [20 points] (b) Suppose instead that just one of the legs is suddenly removed. Find the force exerted by each of the remaining two legs immediately after. 20 points thin solid disk of mass M and radius R, the three principal Possibly useful information: for a MR,MR2, and MR moments of inertia about its center are:

Answer 1

a] Torque about one leg = Mg R

I alpha = Mg R

(1/4 MR^2 + MR^2) * a/R = MgR

5/4 MR a = MgR

a = 4g/5

Force exerted by third leg = Mg - Ma = Mg - Mg*4/5

= Mg/5

b] Let the acceleration just after be a,

Force by each of the two legs is given by equation,

Mg - 2F = Ma

F = M*(g-a)/2

Angular acceleration alpha = a/R cos 60 degree = 2a/R

torque = I alpha

2* F *R cos 60 degree = 1/4 MR^2 *2a/R

M(g-a) R /2 = 1/4 MR^2 * 2a/R

g -a = a

a = g/2

Force exerted by each of the two legs,

F = Mg/4