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In: Physics

LAB QUESTION A rotating cylindrical rod of mass m=4 Kg and length l=1 meter is connected...

LAB QUESTION

A rotating cylindrical rod of mass m=4 Kg and length l=1 meter is connected to a falling weight of 300 grams with a 2 meters string (Maximum distance covered by the falling weight). The distance from the wound string to the axis of rotation is 300 mm. The time taken by the falling weight to halfway is 2 seconds and to the bottom is 3.7 seconds.

Change of angular momentum in the spinning rod

Mass (Kg)

Moment of Inertial of Rod

I

(Kg.m2)

Angular Momentum at halfway

(Kg. m2)/s

Angular Momentum at bottom

(Kg. m2)/s

Difference

(Kg. m2)/s

0.3

  1. Calculate the angular momentum of the rod at both the points (halfway and bottom).
  2. Explain the difference in the angular momentum at two points.

Solutions

Expert Solution

Let the initial velocity of the falling weight be u, then by using the kinematic equation by assuming that the acceleration a is constant,

S = ut + 1/2at2

Now as we know that for halfway and at bottom,

1m = u x 2s + 1/2 x a x 4s2

=> 2u + 2a = 1 ............(1)

And,

2m = u x 3.7s + 1/2 x a x 3.72

=> 2u + 3.7a = 4/3.7 ..........(2)

Solving both equations we get,

a = 0.0477 m/s2

u = 0.4523 m/s

Now, as the string is not slipping on the rod we have

v = r, is angular velocity

Now the velocities at halfway and bottom is given as,

vhalf = u + at

=> vhalf = 0.4523m/s + 0.0477 x 2s = 0.5477 m/s

vbottom = u + at

=> vbottom = 0.4523m/s + 0.0477 x 3.7s = 0.6288 m/s

Thus we have angular velocities as,

= 0.5477 m/s / 300mm = 1.8257 rad/s

= 0.6288 m/s / 300mm = 2.096 rad/s

Now for a rod rotating about its axis we have angular momentum, L = I x , where I is the moment of inertia about that axis,

Now moment of inertia of a cylinder about its axis is given as,

I = 1/2 mR2 = 1/2 x 4kg x (300mm)2 = 0.18 kg m2

Lhalf = 0.18 x ( 1.8257 ) = 0.328626 kg m2 /s2

Lbottom = 0.18 x (2.096) = 0.37728 kg m2/s2

Difference = 0.048654 kg m2/s

This difference is due to the acting torque on the cylinder as torque is also equal to the change in angular momentum per unit time. Thus torque causes change in angular momentum.


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