# Angular acceleration on a pulley

• df102015
In summary, the mass will be rotating at a rate of 0.5 revolutions per minute when the weight has fallen 7.4 meters and was released from rest.

## Homework Statement

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A mass of 6.1 kg tied to a string is wrapped around a disk as shown. If the disk has a mass of 8.2 kg and a radius of 2.3 m, how fast will the disk be rotating when the weight has fallen 7.4 m and was released from rest?

Θ = S/R = x/R
Θ = 0.5 α t^2
τ = I α
τ = T R
T = mg - ma
I = 0.5 M R^2

## The Attempt at a Solution

These are the equations that my teacher gave me. So far, when i combined them i get this...

α = τ/I = T R / 0.5 M R^2 = (mg - ma) R / 0.5 M R^2 = (mg - ma) / 0.5 M R

Which, when numbers are substituted in becomes...

(6.1 x 9.81 - 6.1a) / 0.5 x 8.2 x 2.3 = α

but i still do not have acceleration (a) and the equation
a = Δv / t
gives me two other things that i do not have... change in velocity and time. And the equations using theta that he gave me seem almost useless because the value of theta is not given, although it looks as if it is 90 degrees, but i could be wrong. When i calculate theta using x/R i got 3.2174 which i assume is in pi radians, but could just be wrong altogether. Can somebody please just point me in the right direction, help me with some different equations if they are needed, it is the last problem that i need to finish on the homework. Also i believe my teacher may have mentioned there is a way to solve by using energy equations, if that is easier then please help using that. Thanks so much!

Energy is a nice way to solve it. But you've already done most of the work in using the method of forces and torques. All you need is another relation between α and a. I bet you've covered that relation. It's similar to the equation s = Rθ.

is the equation i am looking for α = a/R ?

Yes, that's it.

df102015
TSny said:
Yes, that's it.
Thanks so much! You have no idea how much i appreciate it!

Another way to solve is by conservation of energy:-

loss in G.P.E. of box = gain in K.E. of box + gain in K.E. of Disc.

## mgh=\frac{1}{2} mv^2 + \frac{1}{2} I\omega^2\ ##
## \omega\ ## is clockwise
and for no slipping ## v=r \omega\ ##

## What is angular acceleration on a pulley?

Angular acceleration on a pulley is the rate of change of angular velocity of a pulley. It is a measure of how quickly the rotational speed of the pulley is changing.

## How is angular acceleration on a pulley calculated?

Angular acceleration on a pulley can be calculated by dividing the change in angular velocity by the change in time. It is expressed in radians per second squared (rad/s²).

## What factors affect angular acceleration on a pulley?

The factors that affect angular acceleration on a pulley include the mass of the pulley, the radius of the pulley, and the torque applied to the pulley.

## What is the relationship between angular acceleration and linear acceleration?

Angular acceleration and linear acceleration are related by the radius of the pulley. The linear acceleration of a point on the edge of a pulley is equal to the angular acceleration multiplied by the radius of the pulley.

## Why is angular acceleration on a pulley important?

Angular acceleration on a pulley is important because it helps us understand the motion of objects connected to the pulley and can be used to calculate the forces and energy involved in the system.