2)Now the right string is cut! What is the initial angular acceleration of the m

Question

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)

3)What is the tension in the left string right after the right string is cut?

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?

5)What is the acceleration of the center of mass of the meterstick when it is vertical?

6)What is the tension in the string when the meterstick is vertical?

7)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

Explanation / Answer

1) m*g/2 = 1.2152 N
2) Net torque about pivot point = m*g*0.25 =0.6076 N-m
Net torque = I .

I = m*l^2/12 + m*(0.25)^2

= 0.0362 kgm2

= 0.6076/0.0362

= 16.8rad/s2

3) acceleration of centre of rod =a = 0.25 * = 4.2rad/s

Net force = m*a

m*g - T =  m*a

T = 1.388 N

4) Work done by gravity = change in rotational kinetic energy about pivot

=> m*g*0.25 = (1/2)* I * ^2

=> = 5.8 rad/s

5) acceleration of COM = Acom = centripetal acceleration as there is no tangential force

= r*^2

= 8.4 m/s2

6) T - m*g = m*Acom

T = 4.51N

7) When the string is cut(when its horizontal). This is because torque is maximum at this time.

(since Net torque = I .)

Hope this helps.

CHEERS!!

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