The Student Room Group

Work Done by a Wheel

https://isaacphysics.org/questions/wheel_and_wall_num?board=5b1c5322-e1a1-4402-a228-191259ecc67a&stage=further_a

I'm a little stuck on what the video means by "final and initial" angle - I thought it would just be theta_0 -> pi/2 or the angular displacement of 0 -> pi/2-theta_0 but none of the methods led to anything.

I tried to derive the expression for the energy as well assuming that at the start and end v = 0, so the translational KE = 0.

Any help greatly appreciated.
Working out:
math.png
Reply 2
Original post by mosaurlodon
https://isaacphysics.org/questions/wheel_and_wall_num?board=5b1c5322-e1a1-4402-a228-191259ecc67a&stage=further_a
I'm a little stuck on what the video means by "final and initial" angle - I thought it would just be theta_0 -> pi/2 or the angular displacement of 0 -> pi/2-theta_0 but none of the methods led to anything.
I tried to derive the expression for the energy as well assuming that at the start and end v = 0, so the translational KE = 0.
Any help greatly appreciated.

Youre correct in saying its integrating from theta0 to pi/2, but you want to find the work done by F (only). Your work done is the right way to go (given the previous hint) and the final part/sentance of the question is a strong hint at what the answer should be.
(edited 1 month ago)
Oh I see...
so the work done by F is just opposing (equal to) the work done from the weight.
So its just Int from theta_0 to pi/2 of rmgcos(theta) dtheta?
And I guess by the question saying "Is this the value you would expect?"
the answer is exactly mgh so the work done by torque is just overcoming (equal to) the GPE?

that got me the right answer but I dont think its right because in the video hint it asks you to calculate F?
(edited 1 month ago)
Reply 4
Original post by mosaurlodon
Oh I see...
so the work done by F is just opposing (equal to) the work done from the weight.
So its just Int from theta_0 to pi/2 of rmgsin(theta) dtheta?
And I guess by the question saying "Is this the value you would expect?"
the answer is exactly mgh so the work done by torque is just overcoming (equal to) the GPE?
that got me the right answer but I dont think its right because in the video hint it asks you to calculate F?

The video hint is really to get an expression for F, not to calculate a value. They do this for a particular value of theta though.

Tbh, the actual maths is relatively simple (as is often the case) and it would be worthwhile thinking about why your KE approach wasnt correct and why F is equal to this expression (is it?). Also could you get an equivalent approach by getting F(x) and integrating over x, where x is the horizontal displacement.
(edited 1 month ago)
Tbh, I still don't really get why you have to work out F(θ) as mgcot(θ) if it is ignored anyways.
So to get the answer you have to do

but if you do that integral for F(θ) you get 22J instead?
https://www.desmos.com/calculator/rlivddmugu

I also tried to get the work done using F(x) but im not really sure how to even get a expression for F(x) so I tried something else - does the following method make sense?
working out.png
Reply 6
Original post by mosaurlodon
Tbh, I still don't really get why you have to work out F(θ) as mgcot(θ) if it is ignored anyways.
So to get the answer you have to dobut if you do that integral for F(θ) you get 22J instead?
https://www.desmos.com/calculator/rlivddmugu
I also tried to get the work done using F(x) but im not really sure how to even get a expression for F(x) so I tried something else - does the following method make sense?working out.png

Youre getting quite a few things mixed up. For the standard approach, by definition, torque or moment is
F(theta)*r*sin(theta)
and in this case F(theta) = mgcot(theta). So integrate that wrt theta between the limits and make sure youre happy with that.

Tbh, as in a previous thread, it hellps to be clearer about writing things up so why would you take sin(theta_0) outside the integral? Why are you treating it as a constant?
(edited 1 month ago)
Ok im good with the F(θ) part now - thank you.

I think the problem in my working was basically saying θ_0 ~ θ (as θ = θ_0 + dθ)
so I just converted all the θ_0s to θs so I can integrate and stop treating them as constants which I think managed to work?
Reply 8
Original post by mosaurlodon
Ok im good with the F(θ) part now - thank you.
I think the problem in my working was basically saying θ_0 ~ θ (as θ = θ_0 + dθ)
so I just converted all the θ_0s to θs so I can integrate and stop treating them as constants which I think managed to work?

As in the previous post, why then does this F(theta) correspond to work done moving the wheel as its the equibrium value?

As for treating it as a constant, you could think of the theta~pi/2 scenario, then F~0 as its perpendicular to the weight. So fairly obviously, it must be a function of theta and hence it is not constant.

The previous post possibly hides the relative simplicity. Youre integratng the torque wrt theta and you really dont need to make any infinitesimal arguments to set it up.
(edited 1 month ago)
"why your KE approach wasnt correct" - I believe my original working wasnt correct because I was basically taking the integral of 0 as I thought it would just be integral of resultant torque.

I think F(θ) corresponds to the work done even though its an equilibrium force because of the normal reaction force?
So initially the mg cancels with the N and the wheel can rotate with a very small F but as soon as the wheel comes off the ground the mg force comes into play and if F is equal to this mg force then there will be no resultant torque, but since the wheel was already moving from the beginning it has some rotational momentum so carries on rotating?
Original post by mosaurlodon
"why your KE approach wasnt correct" - I believe my original working wasnt correct because I was basically taking the integral of 0 as I thought it would just be integral of resultant torque.
I think F(θ) corresponds to the work done even though its an equilibrium force because of the normal reaction force?
So initially the mg cancels with the N and the wheel can rotate with a very small F but as soon as the wheel comes off the ground the mg force comes into play and if F is equal to this mg force then there will be no resultant torque, but since the wheel was already moving from the beginning it has some rotational momentum so carries on rotating?

You could argue that any torque greater than the equilbrium value will cause the wheel to hop over the step. If its exactly equal to the equilbrium value, it wont move and if its much greater then the speed of the wheel will be significant and some of the work done will be converted into KE. So here the assumption is that its epsilon greater than the equilbrium value and the speed when it reaches theta=pi/2 is essentially zero.

Thinking about KE therefore isnt the right approach as youre essentially considering zero (translational) speed and doing work against gravity.

You could revisit more usual (not torque) work done. F is a horizontal force and work done is the integral of that wrt (horizontal displacement) x. So really you just need to get the function f(x) where
cot(theta) = f(x)
where x is the base of the triangle, and do the integral wrt x. Its not that much harder (mathematically) than the polar/torque approach and arguably easier to get set up in the first place.
(edited 1 month ago)
I just want to double check the method of getting work done by integrating F wrt displacement.

So get F(x) by working out cot(theta) = f(x) and then F(x) = mgf(x)
and get f(x) using x = rcos(theta_0) - rcos(theta)?
then integrate F(x) from 0 to rcos(theta_0)?
(edited 1 month ago)
Original post by mosaurlodon
I just want to double check the method of getting work done by integrating F wrt displacement.
So get F(x) by working out cot(theta) = f(x) and then F(x) = mgf(x)
and get f(x) using x = rcos(theta_0) - rcos(theta)?
then integrate F(x) from 0 to rcos(theta_0)?

In a sense this should be the easiest as its the basic definition of work done. You know the limits of x (horizontal displacement so 0 to rcos(theta_0), though its probably easier to just get the latter from pythagoras) and you can get cot(theta) in terms of the base length x as the hypotenuse is 0.5, again pythagoras. Then its just a case of integrating F(x) wrt x over the limits, which is obviously related to trig integrals.
(edited 1 month ago)
Oh nice I see it now - I guess my old method in blue was just some weird twisted way that turned out to work.
thank you so much :smile:

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