MURTUZA ZAVERI
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18 Jul 2017

# converting linear velocity to angular velocity

THE title of this thread might appear to be quite misleading or something, but i did not know how best to put it though.

now here is a situation:

a dead weight weighing 10 kg has been hoisted up to a certain height, by a rope and pulley mechanism. it is then made to remain hanging there, by tying the other end of the rope to some support.

now if all of a sudden, if the other end of the rope was released, causing the weight to fall freely, and dragging the rope with it, through the pulley, then what would be the rpm of the rotating pulley ?

by using the formulas of potential and kinetic energies, the velocity of the falling weight can be determined. now how do i use this to find the rotational speed of the pulley
18 Jul 2017
Assuming that the rope is inelastic, there is no slip between the rope and the pulley, and the pulley is frictionless, the speed of the weight is given by:
V=0.5 g t^2.
RPM = V/(pi X d) d is the pulley diameter
This means that the pulley RPM continuously increases.
Atwood's Machine demonstrates this.
Atwood's Machine
We had this in our high school.
Personally this experiment was the one that made me decide on a research career in 1953.

MURTUZA ZAVERI

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19 Jul 2017
thankyou Ramani sir

MURTUZA ZAVERI

Branch Unspecified
19 Jul 2017
A.V.Ramani
Assuming that the rope is inelastic, there is no slip between the rope and the pulley, and the pulley is frictionless, the speed of the weight is given by:
V=0.5 g t^2.
RPM = V/(pi X d) d is the pulley diameter
This means that the pulley RPM continuously increases.
Atwood's Machine demonstrates this.
Atwood's Machine
We had this in our high school.
Personally this experiment was the one that made me decide on a research career in 1953.
sir,

what about the units of the formulas you just provided.
19 Jul 2017
V - velocity m/sec
t - time from start in seconds
g - acceleration due to gravity = 9.8 m/sec^2
d - diameter in m
Pi - 3.1416

MURTUZA ZAVERI

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20 Jul 2017
thanx again sir