Question

A meterstick (L = 1 m) has a mass of m = 0.128 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

1)

What is the tension in the left string?
N

2)

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)
rad/s²

3)

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

4)

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 angular speed when the meterstick is vertical?
rad/s

5)

What is the acceleration of the center of mass of the meterstick when it is vertical?
m/s²

6)

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

7)

Where is the angular acceleration of the meterstick a maximum?

a) right after the string is cut and the meterstick is still horizontal

b) when the meterstick is vertical – at the bottom of its path

c) the angular acceleration is constant

Answer 1

Т. T 경 Ift T be tension on each string ang mg be weight of meter stick 025-> 0.25 then ng at equî li bonium 2T= mg To mora 2 0.128x9.8 2 Oo 627 N at Tension of in the left string = 8.6270 when ref right stering is cut & DA > mo integra met tonque about pirot point a Mg so 0.128 11908 8x0.25 0.314. N-m IRI be moment of the stick about a then using pamalled anis theorem of imeytia about a T2 moment ineuria about centre of mass t moment about Ma2 (20) M22 md2 Thom ent or Anettia 12 0.128412 10.1983 (6.25; 2=40732 12 I rader I= 0.0184 kg m 2 If a be angwam acceley alfon then we can repate conque about As moment of in extra and ang Wan acceleration = ra 문 olent 0.314 Ono 187 2 16.779! mad's2 acceleration of centre of mood 16.791 0:25 = 4,198 ms 2 force och merey shok ama mg-Tema fequation Hon of motions stien Ta mg-ma => 0:28 (9.8-4.198) N ora 2N Tension in Left string after cutting right string Tao.72N Net up Now work done by gravity Rotation change in eney gy the mod h work dong by goramity o mgh change in Rotational kinete eney gy & Tw2 pco= ang wan) speed of shor mgha / 1 w2 La mgh 'I 2x0:128x19,8 %0.25 Orol87 5.79! vads ang way speed of stick when meney Shek is Ney ficat e 50791 rady When stren vertical aceefesyoutoon of centre of mass a com a centripetal acceleyalton as their is no tangential force when sffek is neuffcal ass w2 5079120125 MAR 8. 384 mg 6 AT when shick is veftical then To ng = ma com Ta mgtm acom s 01289.8 +8.384)^ 2.33 N Tension of stick when it is neutical T= 2.33N when the string is cut and meter stick is horizontal Tonque is maximum at this time Torque To Ial so angway acceleration of merey stiera is maxran um when stick is right after the storing is cut and meter stick is still honizon fal, since Scanned with CamScanner