The Student Room Group

CYLINDER SLIDING DOWN A SLOPE physics question

So my working for this question is https://isaacphysics.org/questions/cylinder_sliding_slope?board=a425eb19-5203-4fb5-87e9-c10ba8e6d5e9&stage=all

I think I have to take moments, so the friction force, F_r and the T. The component of the weight acting along the slope I believe cancels out as I am taking moments about the centre? So when taking moments F_r*D/2=T*D/2*sin(alpha). However alpha I am unsure of. The height at which the Tension is acting upon is h. I know that T depends on h, since more h would mean more torque from T as the angle decreases to 90º and less h would mean less torque from T as the angles increases to 180º. So should I use trigonometry? then I would have a right-angle triangle, where the hypothenuse is the radius, the vertical component is therefore h-D/2 and the horizontal component sqrt((h-D/2)^2+D^2/4). However I am unsure as to how approach the situation. Please let me know wether I am totally off track which is highly possible.
Original post by Javier García
So my working for this question is https://isaacphysics.org/questions/cylinder_sliding_slope?board=a425eb19-5203-4fb5-87e9-c10ba8e6d5e9&stage=all
I think I have to take moments, so the friction force, F_r and the T. The component of the weight acting along the slope I believe cancels out as I am taking moments about the centre? So when taking moments F_r*D/2=T*D/2*sin(alpha). However alpha I am unsure of. The height at which the Tension is acting upon is h. I know that T depends on h, since more h would mean more torque from T as the angle decreases to 90º and less h would mean less torque from T as the angles increases to 180º. So should I use trigonometry? then I would have a right-angle triangle, where the hypothenuse is the radius, the vertical component is therefore h-D/2 and the horizontal component sqrt((h-D/2)^2+D^2/4). However I am unsure as to how approach the situation. Please let me know wether I am totally off track which is highly possible.

@mqb2766 🙂
Reply 2
Original post by Javier García

Ive not tried too hard to decipher your approach - an uploaded diagram helps, but you dont know h or T, so it would make sense to take moments about where T intersects the diameter (height h above the plane along the diameter) so there are no moments assoicated with T (and N).

Then its just a case of balancing the moments due to friction and weight which is a bit of simple trig.
Original post by mqb2766
Ive not tried too hard to decipher your approach - an uploaded diagram helps, but you dont know h or T, so it would make sense to take moments about where T intersects the diameter (height h above the plane along the diameter) so there are no moments assoicated with T (and N).
Then its just a case of balancing the moments due to friction and weight which is a bit of simple trig.

I understand it is quite tedious to read through my working :frown:. However, I will try to upload it. Nevertheless, you said that there are no moments due to the Tension? How come that be possible? Also, the weight acting along the inclined plane acts about the centre of the circumference so shouldn't it be zero? hence I should equate the moment due to friction and the tension?
Reply 4
Original post by Javier García
I understand it is quite tedious to read through my working :frown:. However, I will try to upload it. Nevertheless, you said that there are no moments due to the Tension? How come that be possible? Also, the weight acting along the inclined plane acts about the centre of the circumference so shouldn't it be zero? hence I should equate the moment due to friction and the tension?

You take moments about the point where the tension intersects with N (well the arrows / line of action associated with them). Both are then zero perpendicular distance from where you take moments so they have zero moment.

Tbh, choosing where you take moments is one of the things to keep things simple. You know you want to involve mu, so the only other two possibilities are the centre (eliminate W) or the previous point (eliminate T). As W will be involved in friction and you dont know T, then the choice is fairly obvious.
(edited 2 months ago)
Original post by mqb2766
You take moments about the point where the tension intersects with N (well the arrows / line of action associated with them). Both are then zero perpendicular distance from where you take moments so they have zero moment.
Tbh, choosing where you take moments is one of the things to keep things simple. You know you want to involve mu, so the only other two possibilities are the centre (eliminate W) or the previous point (eliminate T). As W will be involved in friction and you dont know T, then the choice is fairly obvious.

Thanks! I got the answer, it was a sound idea to take moments about the point of intersection between T and N, hence you eliminate those as there are acting at 0º. Then you consider the weight and the friction.
Reply 6
Original post by Javier García
Thanks! I got the answer, it was a sound idea to take moments about the point of intersection between T and N, hence you eliminate those as there are acting at 0º. Then you consider the weight and the friction.

As above, its one of the most important things to do. Otherwise youll have to get two simultaneous equations in T and h and then eliminate T and solve for h. When you eliminate T, youll end up with "the same" equation as if youd just taken moments about this point in the first place. So a lot of work for nothing.
Original post by mqb2766
As above, its one of the most important things to do. Otherwise youll have to get two simultaneous equations in T and h and then eliminate T and solve for h. When you eliminate T, youll end up with "the same" equation as if youd just taken moments about this point in the first place. So a lot of work for nothing.

Yes, however I don't think I would know how to approach it that way, as the angle the tension T makes with the radius would be unbeknownst, it would depend on h but I would totally be lost. So, would you know how to do it that way? Or is it a bit complicated since we don't know that angle? Finally, it is paramount to take moments about a point that simplifies things, I didn't know you could take moments inside the circunference so thanks a lot.
Reply 8
Original post by Javier García
Yes, however I don't think I would know how to approach it that way, as the angle the tension T makes with the radius would be unbeknownst, it would depend on h but I would totally be lost. So, would you know how to do it that way? Or is it a bit complicated since we don't know that angle? Finally, it is paramount to take moments about a point that simplifies things, I didn't know you could take moments inside the circunference so thanks a lot.

If you dont take moments about a point that "simplifies things" then youll have to get 2 (or more) simultaneous equations and have to eliminate the unknowns youre not interested in and ... Error prone and long winded but its not wrong.

You can take moments about anywhere. So if you had a ladder against a wall, you could take moments from your toilet, hypothetically speaking. If the object isnt rotating, then its not rotating about any point.

Not sure which radius youre talking about (first part of your response). As in the previous posts, it would help to see a picture of what youre talking about.
(edited 2 months ago)
Original post by mqb2766
If you dont take moments about a point that "simplifies things" then youll have to get 2 (or more) simultaneous equations and have to eliminate the unknowns youre not interested in and ... Error prone and long winded but its not wrong.
You can take moments about anywhere. So if you had a ladder against a wall, you could take moments from your toilet, hypothetically speaking. If the object isnt rotating, then its not rotating about any point.
Not sure which radius youre talking about (first part of your response). As in the previous posts, it would help to see a picture of what youre talking about.
I do not know why it is not letting me upload photographs, but what I meant by radius is that, to my understanding, the moment due to the tension about the centre would be T*D/2*sin(alpha), where D/2 is the radius as that is the distance to which the T is acting with respect to the centre right? and then the sin(alpha) is the angle it makes with the line that connects the centre with the tension which would be the radius, so if you have a rope attached with glue to the surface of the cylinder, then the line connecting that point of attachment to the centre would be the radius. Also, you said "if the object is not rotating" is it that I can only take moments about any point provided it is not rotating, so when it is rotating I can only take moments about the point of rotation?
Reply 10
Original post by Javier García
I do not know why it is not letting me upload photographs, but what I meant by radius is that, to my understanding, the moment due to the tension about the centre would be T*D/2*sin(alpha), where D/2 is the radius as that is the distance to which the T is acting with respect to the centre right? and then the sin(alpha) is the angle it makes with the line that connects the centre with the tension which would be the radius, so if you have a rope attached with glue to the surface of the cylinder, then the line connecting that point of attachment to the centre would be the radius. Also, you said "if the object is not rotating" is it that I can only take moments about any point provided it is not rotating, so when it is rotating I can only take moments about the point of rotation?

If you took moments about the centre, then the perpendicular distance of T from it would be (h-D/2) so the moment would be
T(h-D/2)
What you say isnt wrong, but no need to introduce alpha and sin(alpha) would be
(h-D/2)/(D/2)
and is a bit of a circular argument ...

Im not sure how much you do on rotating dynamics. A moment is also known as a torque so the rotational force and if the object isnt in equilibrium then it youll equate the net torque (moment) to the angular acceleration multiplied by the moment of inertia, so the rotational equivalence of newton 2 or f=ma.

Quick Reply