Question

t. A meter‐stick of negligible mass is pivoted in the middle so it can spin horizontally. Two 500g clamps (consider point masses) are placed at the 40cm and 60cm marks. A) Find the moment of inertia of the system. B) One of the clamps is pushed, at an angle of 42˚ with respect to the meter‐stick, with a force of 12N for a time period of 0.2s. What will be the meter stick’s angular velocity? C) As this meter‐stick is spinning, the clamps slip over to the 30cm and the 90cm marks respectively. What will be meter stick's angular velocity?(A=.01kgm²; B=16.06rad/s; C=1.61rad/s)

Answer 1

A) let

m1 = m2 = 500 grams

= 0.5 kg

moment of inertia of the system, I = m1*r1^2 + m2*r2^2

= 0.5*0.1^2 + 0.5*0.1^2

= 0.01 kg.m^2

B) Torque T = r cross F

= r*F*sin(theta)

= 0.1*12*sin(42)

= 0.803 N.m

angular acceleration, alfa = T/I

= 0.803/0.01

= 80.3 rad/s^2

now use, w = wo + alfa*t

= 0 + 80.3*0.2

= 16.06 rad/s

C) In this case one clam is at 20 cm from the center ans another clamp is at 40 cm from the center.

If = 0.5*0.2^2 + 0.5*0.4^2

= 0.1 kg.m^2

apply conservation of angular momentum

If*wf = Ii*wi

wf = Ii*wi/If

= 0.01*16.06/0.1

= 1.606 rad/s