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) Anwser (16.8)

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?

Explanation / Answer

1) As the meter stick is in equilirium, net force and net torque acting on it must be zero.

Apply, Net torque about right string = 0

FL*0.5 - m*g*0.25 = 0

FL = 0.25*m*g/0.5

= 0.25*0.15*9.8/0.5

= 0.735 N

2) moment of Inertia ofstick about left string,

I = m*L^2/12 + m*d^2

= 0.15*1^2/12 + 0.15*0.25^2

= 0.021875 kg.m^2
Torque acting about left string,

Torque = m*g*0.25

= 0.15*9.8*0.25

= 0.3675 N.m

angular acceleration, alfa = T/I

= 0.3675/0.021875

= 16.8 rad/s^2

3) FL = m*g

= 0.15*9.8

= 1.47 N

4) Apply conservation of energy

final kinetic energy = initial potentail energy

0.5*I*w^2 = m*g*0.25

0.5*0.021875*w^2 = 0.15*9.8*0.25

w^2 = 0.3675/0.0109375

w = sqrt(0.3675/0.0109375)

= 5.8 rad/s

5) a_rad = r*w^2

= 0.25*5.8^2

= 8.41 m/s^2

6) T = m*g + m*a_rad

= 0.15*9.8 + 0.15*8.41

= 2.7315 N

Related Questions

Navigate

• Browse (All)
• Browse 2
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.