Calculating Angular Acceleration and Speed with Variable Force on a Pulley

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Homework Help Overview

The problem involves calculating the angular acceleration and speed of a pulley subjected to a time-varying tangential force. The pulley has a specified rotational inertia and radius, and the force applied at its rim is defined as a function of time.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the implications of a time-varying force on angular acceleration and torque. There is an exploration of the relationship between torque and angular acceleration, with some suggesting the need for integration to find the change in angular velocity.

Discussion Status

Participants are actively engaging with the problem, questioning assumptions about the relationship between torque and angular acceleration. Some have provided guidance on the need to consider the functional form of angular acceleration and how it relates to angular velocity over time.

Contextual Notes

There is a mention of significant figures in the context of the final answer, indicating a consideration of precision in calculations. The original poster expresses confusion about the discrepancy between their calculated result and the expected answer.

Arman777
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Homework Statement


A pulley with a rotational inertia of ##10^{-3}\frac {kg} {m^2}## about its axle and a radius of ##10 cm##, is acted on by a force applied tangentially at its rim.The force magnitude varies with time as
##F=0.5t+0.3t^2## , with ##F## in Newtons and ##t## in seconds.The pulley is initially at rest At ##t=3.0s## what are its (a) angular acceleration (b) angular speed ?

Homework Equations


##τ(t)=F(t)rsinθ##
##w-w_0=∝(t)Δt##
##τ(t)=I∝(t)##

The Attempt at a Solution


The confusing part for me is that F varies with time so from that angular acceleration will vary with time also torque will vary with time.
##τ(t)=F(t)rsinθ## ,
##sinθ## is always ##1## so ##τ(t)=F(t)r## is our equation.
And we also know that ##τ(t)=I∝(t)##
so
##F(t)r=I∝(t)## then ,
##∝(t)=\frac {F(t)r} {I}##
##∝(t)=420\frac {rad} {s^2}##

for B I wrote,
##w-w_0=∝(t)Δt## the object is initally at rest so ##w_0=0## then ##w=420\frac {rad} {s^2}.3s## which its
##w=1260\frac {rad} {s}##

but answer says its ##w=500\frac {rad} {s}##
where I am going wrong ?

Thanks
 
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Arman777 said:
∝(t)=420rads2
This is where you went wrong. The acceleration is a function of time because the torque is a function of time, you said so yourself. Look at the equation above this one and put in the correct functional form for the force. To find the correct change in angular velocity, you need to do an integral.
 
##τ=\int_0^3 F(t)r \, dt## ?
 
Arman777 said:
##τ=\int_0^3 F(t)r \, dt## ?
This expression does not give the torque. Your original expression ##\tau (t)=F(t) r## correctly gives the torque at any time t.

Using the expression for the torque can you find the angular acceleration ##\alpha (t)##? If you know ##\alpha (t)##, what do you have to do to find ##\omega (t)##? How are the two related in general not just when the acceleration is constant?
 
##ω_f-ω_i=\int ∝(t)\, dt##
 
That's it. Now find the correct α(t).
 
I found ##495\frac {rad} {s}## answer is ##500\frac {rad} {s}## ? Is it ok ?
 
How many significant figures do you think you should carry? Look at the significant figures of the numbers that are given to you.
 
oh I see ok
It will be ##4,95.10^{2}\frac {rad} {s}=5.10^{2}\frac {rad} {s}##
 
  • #10
Thanks
 

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