Calculating Angular Acceleration and Speed with Variable Force on a Pulley

AI Thread Summary
The discussion revolves around calculating the angular acceleration and speed of a pulley subjected to a time-varying tangential force. The force is defined as F=0.5t+0.3t^2, leading to a torque that also varies with time. The initial attempt at solving the problem incorrectly assumed constant angular acceleration, resulting in an erroneous angular speed calculation. The correct approach involves integrating the angular acceleration over time to find the change in angular velocity. Ultimately, the correct angular speed at t=3.0s is approximately 500 rad/s, emphasizing the importance of significant figures in the final 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}##
 
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Thanks
 
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