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 2.7 seconds.

If we replace the 300-gram falling weight with the 2 Kg weight, it takes 1 seconds to reach to the halfway mark and 1.5 seconds to the bottom.

*Change of angular momentum in the spinning rod*

Mass (Kg) |
Angular Momentum at halfway (Kg. m |
Angular Momentum at bottom (Kg. m |
Difference (Kg. m |

2 |

Calculate the corresponding angular momentum at halfway and bottom

Answer #1

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 2.7 seconds.
Change of angular momentum in the spinning rod
Mass (Kg)
Moment of Inertial of...

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.
!!!! If we replace the 300-gram falling weight with the 2 Kg
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