Calculating Angular Acceleration and Speed with Variable Force on a Pulley

In summary, a pulley with a rotational inertia of ##10^{-3}\frac {kg} {m^2}## and a radius of ##10 cm## is acted on by a force that varies with time. At ##t=3.0s## the pulley has an angular acceleration of ##420\frac {rad} {s^2}## and an angular speed of ##500\frac {rad} {s}##. The correct answer for the angular speed is ##5.10^{2}\frac {rad} {s}## due to significant figures.
  • #1
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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  • #2
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.
 
  • #3
##τ=\int_0^3 F(t)r \, dt## ?
 
  • #4
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?
 
  • #5
##ω_f-ω_i=\int ∝(t)\, dt##
 
  • #6
That's it. Now find the correct α(t).
 
  • #7
I found ##495\frac {rad} {s}## answer is ##500\frac {rad} {s}## ? Is it ok ?
 
  • #8
How many significant figures do you think you should carry? Look at the significant figures of the numbers that are given to you.
 
  • #9
oh I see ok
It will be ##4,95.10^{2}\frac {rad} {s}=5.10^{2}\frac {rad} {s}##
 
  • #10
Thanks
 

What is torque with variable force?

Torque with variable force is a measure of the rotational force applied to an object, taking into account the changing force vectors over time.

How is torque with variable force calculated?

To calculate torque with variable force, you must integrate the force function with respect to the displacement function. The resulting integral will give you the total torque applied over the given displacement.

What factors affect torque with variable force?

The main factors that affect torque with variable force are the magnitude and direction of the force vectors, as well as the distance from the point of rotation at which the force is applied.

What is the difference between torque with constant force and torque with variable force?

The main difference between torque with constant force and torque with variable force is that with constant force, the force applied remains the same throughout the rotation, while with variable force, the force changes over time.

Why is torque with variable force important in physics and engineering?

Torque with variable force is important in physics and engineering because it allows us to understand and predict the rotational motion of objects under the influence of changing forces. This is crucial in designing and building machines and structures that require precise control of rotational forces.

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