# Moments and forces, when part of the scenario involves a circle or arc rather than a straight line

Some questions below, involving strategies, regarding the following problem:

A hoop of weight $20$ Newtons can rotate freely about a pin fixed in a wall. A string has one end attached to a pin, runs round the circumference of the hoop to its lowest point, and is then held horizontally at its other end. A gradually increasing horizontal force is now applied to the string, so that the hoop begins to rotate about the pin. Find the tension in the string when the hoop has rotated through $40$ $degrees$.

1. Even though there is circular motion here, with the string at least, - can this be modelled with triangulation between $W$, $T$ and one point of the hoop i.e. make a right angled triangle for which moments can be worked out?

For instance $W$ = $20 Newtons$ initially and then $W$ = $20cos(40)$ after movement.

1. Would this be accurate enough or would the fact that part of the "path" of the string is circular even though no specfic information about the radius or circumference is given? A lot of scenarios involve ladders or rods that are straight.
• Does the hoop rotate freely around its centre point? – Gert Aug 19 '15 at 23:30
• A diagram would help. – paisanco Aug 20 '15 at 2:47
• No, the hoop can only rotate around the pin that is fixed on the wall. This is the same pin that the string starts from. The 40 degrees rotation, when the string is pulled to the left, is from the vertical axis of the pin, if you drew a vertical and horizontal axis where the pin is located. – J132 Aug 20 '15 at 10:12 Definitions and assumption:

Mass of the hoop $m$, radius of the hoop $R$, force exerted by the string $F$, angle of rotation $\theta$ ("theta"), pivotal point P. It is assumed the string fits around the hoop without friction.

Inertial Moment of the hoop:

Around its centre axis the moment of inertia is $I_c=mR^2$ and with the Parallel Axis Theorem the moment of inertia around the pivot P is $I=2mR^2$.

1. At $\theta=0$:

At $\theta=0$ the only moment provided is $2RF$ so with $\omega$ the angular velocity $\omega =\frac{d \theta}{dt}$:

$2RF=I\frac{d \omega}{dt}$, so $F=mR \frac{d \omega}{dt}$. But this applies only when $\theta = 0$, so we just can't integrate this DE.

1. At $\theta > 0$:

At $\theta > 0$ an extra moment is introduced from the $mg \sin \theta$ component of the weight $mg$, namely the moment $(mg \sin \theta) R$.

The moment provided by $F$ also changes to $F|AP|$, so that the angular equation of motion becomes:

$F|AP|= I \frac{d \omega}{dt} + (mg \sin \theta) R$.

I see no easy way of expressing $|AP|$ as a function of $\theta$ (and other knowns). It would appear $F$ increases as $\theta$ increases because $|AP|$ decreases and $\sin \theta$ increases, certainly for a static limit ($\frac{d \omega}{dt}=0$).

It would be easier if the force $F$ was always perpendicular to the diameter of the hoop, as the moment provided by $F$ would then be a constant $2RF$.

I do not think this problem can be solved at all by the weight of the hoop alone, without taking into account its radius $R$.

• Thanks for the reply. Yes, I thought that the radius was needed. There is a similar question that I have seen that gives the radius of the hoop as 0.5m and a mass of 2Kg and an angle of 30 degrees ($F$=2x9.8xsin30x0.5=$4.9N$), although that is for a rough peg and the tension $F$ is measured from the diameter of the hoop (opposite the peg), i.e. perpendicular to the diameter of the hoop. With the problem we are considering, $F$ is measured from the base of the hoop, exactly as the diagram shows. The answer is given as $7.28N$. – J132 Aug 21 '15 at 8:12 Take moments about $o$.

Radius drawn up to the point of contact is always perpendicular to the tangent.

• This site uses Mathjax for mathematics formatting. Please use that notation, instead of scanned notes, for mathematics. – Emilio Pisanty Nov 29 '17 at 20:14

Gert's answer: AP=cos $\theta$ R+R And moment of inertia is also depending on R.