Question

A meterstick (L = 1 m) has a mass of m = 0.175 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

1)Now the right string is cut! What is the initial angular acceleration of the meterstick about its pivot point? (You may assume the rod pivots about the left string,and the string remains vertical)
2)What is the tension in the left string right after the right string is cut?
3)After the right string is cut, the meterstick swings down to where it is vertical for an instant before it swings back up in the other direction. What is the angularspeed when the meterstick is vertical?

Answer 1

from the figure we can see,


when the right string is cut,

angular acceleration about pivot point on the left,

= mg*.25/(ml^2/12 + m*.25^2) = 16.8 rad/s2 <--------ans(1)

so the acceleration of the centre of mass = 16.8*.25 = 4.2 m/s2
also acceleration of the centre of mass= (T-mg)/m= T/m -g
so equating, we get,
T/m =4.2+9.8 =14
so, T= 14*.175 =2.45N
tension in the string= 2.45N <---------ans(2)

conserving energy,

mg*(.25)= .5*m*v^2 + .5*I*ω^2 = .5*mω^2*.25^2+ .5*(ml^2/12 + m*.25^2)*ω^2

so, ω = 4.85 rad/s<-----------ans(3)

Answer 2

from the figure we can see,

2*T = T1+T2 = mg =1.715

so,T1=T2=T = 0.858N

when the right string is cut,

angular acceleration about pivot point on the left, = mg*.25/(ml^2/12 + .5*m*.25^2) = 21.4 rad/s <---ans(1)

tension in the string= 0.858N <---ans(2)

conserving energy,

mg*(.25)= .5*m*v^2 + .5*I*ω^2 = .5*mω^*.25^2+ .5*(ml^2/12 + .5*m*.25^2)*ω^2

so, ω = 5.3 rad/s