A meterstick (L = 1 m) has a mass of m = 0.21 kg. Initially it hangs from two sh

ID: 1454712 • Letter: A

Question

A meterstick (L = 1 m) has a mass of m = 0.21 kg. Initially it hangs from two short strings: one at the 25 cm mark and one at the 75 cm mark.

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.

1.What is the angular speed when the meterstick is vertical?

2.What is the acceleration of the center of mass of the meterstick when it is vertical?

3.What is the tension in the string when the meterstick is vertical?

In this situation the center of mass comes 25 cm down from the initial orientation.

So the work done due to gravity force = mgh = .21*9.81*0.25 = 0.52 J

Work done by other forces (tension) is zero.

Suppose at the position when stick is vertical the angular velocity is omega.

Calculating the moment of Inertia about the point connected to left rope,

I = (3m/4)*(3L/4)^2/3+(m/4)*(L/4)^2/3 = (3*0.21/4)*(3*1/4)^2/3+(0.21/4)*(1/4)^2/3 = 0.031 kg m^2

Applying work energy concept here, we have, work done = change in kinetic energy

0.52 = 0.5 I omega^2 = 0.5*0.031*omega^2

omega = 5.83 rad/s

2) When it is vertical the torque as well as net force is zero hence there won't be any acceleration at this point.

3) Since there in no net force the tension will be equal to the gravity force,

T = 0.21*9.81 = 2.06 N

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