Question

The next eight questions pertain to the situation described below. A meterstick (L-1 m) has a mass of m=0.224 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? 2Now 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 sa 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 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? "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 h, when the meterstick is vertical-at the bottom of its path the angular acceleration is constant Relow is some snace to write notes on this nroblem

Answer 1

1)

Tension, T = mg/2 = 0.224 x 9.8/2 = 1.098 N

2)

Net torque, T = 0.224 x 9.8 x 0.25 = 0.549 Nm

Moment of inertia, I = (0.224 x 1/12) + (0.224 x 0.25^2) = 0.0327 kg m^2

Angular acceleration, a = 0.549/0.0327 = 16.81 rad/sec^2

3)

Acceleration, a = 0.25 x 16.81 = 4.20 m/s^2

Tension, T = 0.224 (9.8 - 4.2) = 1.25 N

4)

mg x 0.25 = 0.5 I ω^2

(0.224 x 9.8 x 0.25) = 0.5 x 0.0327 ω^2

ω = 5.79 rad/sec

5)

Acceleration, a = 5.79^2 x 0.25 = 8.38 m/s^2

6)

Tension, T = (0.224 x 8.38) + (0.224 x 9.8) = 4.07 N

7)

Option (a)

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