Question

A meterstick (L = 1 m) has a mass of m = 0.233 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

*What is the tension in the left string right after the right string is cut?

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 acceleration of the center of mass of the meterstick when it is vertical?

*What is the tension in the string when the meterstick is vertical?

*Where is the angular acceleration of the meterstick a maximum?

_right after the string is cut and the meterstick is still horizontal

_when the meterstick is vertical – at the bottom of its path

_the angular acceleration is constant

Answer 1

» Isong/2 2) = 0.233(447) - 1.141 acceleration of com is arw 2 - my (0-25) - 0.233(9. 0.233(9.8) 10.25) acceleration of a -0.57085 Nm Toml?mo-25) 0-233C10-273(0-25) 20.0766 kg my 2= 7/- 0.57085 27.452 rod/seet 0.0166 mlfein 0.25(7.452) = 3.726 a=0.250 Netforce = ma nget ma megra) - T T = 6.233 [9.83.726] c 1.4152N ng (0.25) = 1/₂ Iw (0.233 x 9.8 x 045 2 (0.0766) wr w = 0.233 *9.880.25% 0.5800766 W = 3.860 sud/fec. a com ro - 3.4249 m/seer. T-mg a macom T= mcg + Acom) = 0.233 [9.8+3.4249] - 3.15130N when the string is cut (when its horizontal) becaus torque is mariam at this time. u