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How would I calculate the amount of torque required for a motor to move the bar in the diagram below? I'm looking the general solution to solve this. The givens are below. The drawing is of a cart that is moving to the right at a velocity of $v$.

The rope is wrapped around the pulley once before connecting to the fixed point of the bar.

Assume the bar has a mass of $m$, the motor has a pulley with a radius of $r_\mathrm{pulley}$ attached to it, and the rope tie points are equidistant to pivot points, this distance is $r_\mathrm{fixed}$. Also there is friction acting on the bar.

Force diagram

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  • $\begingroup$ Is there something resisting the motion, like friction? $\endgroup$ – ja72 May 31 '15 at 22:26
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This cart is an example of really bad engineering. While the rope looks fixed, any change in angle will make it lose contact with the motor. When you turn the bar clockwise, the top part of the rope does not shrink at the same rate as the lower part, making the rope loose.

However, if you still want to find the torque, you can constrain the rope to stay attached to the motor axle. Then there will be tension in only one side of the rope, while the other gets loose and exerts no force on the bar.

In this way you can find the torque on the bar with the usual formula*:$$\vec\tau=\vec r\times \vec f$$

In order to accelerate the bar, this value must be greater than the torque by friction. Since the rope comes out of the axle perpendicular to the radial direction, the torque on the motor is just the product of the radius with the force (tension on the rope).

The proof that the rope is not stretched is pretty interesting mathematically.

By pythagoras theorem, the length of the rope before turning the bar is:$$\sqrt{r^2+d^2} + \sqrt{r^2+d^2}=2\sqrt{r^2+d^2}$$

Here I use $r$ as the distance from the axis to the tie points, and $d$ as the distance from the axis to the motor.

Now we draw a circle around the bar. When it rotates clockwise by an angle of $\theta$, the $x$ coordinate of the upper tie increases by $r\sin\theta$, while the bottom shrinks by the same amount. Meanwhile, the $y$ coordinates both shrink by a factor of $\cos\theta$, giving a total rope lenght of:

$$L=\sqrt{{(r\cos\theta)^2}+(d-r\sin\theta)^2}+\sqrt{{(r\cos\theta)^2}+(d+r\sin\theta)^2}= \sqrt{r^2\cos^2\theta+r^2\sin^2\theta+d^2-2dr\sin\theta}+\sqrt{r^2\cos^2\theta+r^2\sin^2\theta+d^2+2dr\sin\theta}=\sqrt{r^2+d^2-2dr\sin\theta}+\sqrt{r^2+d^2+2dr\sin\theta}$$ Comparing this with the first equation, you may notice that their values coincide when $\theta$ equals zero. But do they coincide at other values? To answer this, a useful inequality comes in hand. The square root is a concave function, which means that it's graph is always curving down.

Because of this, the sum of distinct values is always smaller than what would be expected with linear functions:

Imagine that the value of $\sqrt{x-dx}$ is smaller than $\sqrt{x}$ by an amount $e$. The value at $\sqrt{x+dx}$ would not be greater than $\sqrt x$ by the same amount $e$, as would be expected in a linear function, but a little bit less than that, because the function grows more slowly for greater values.

Their sum, $\sqrt{x-e}+\sqrt{x+e}$ would, therefore, be smaller than $(\sqrt{x}-e)+(\sqrt{x}+e)=2\sqrt x$. Convex functions have the inequality the other way around, since they grow faster with time.

In the same way, you can say that: $$\sqrt{r^2+d^2-2dr\sin\theta}+\sqrt{r^2+d^2+2dr\sin\theta}\lt2\sqrt{r^2+d^2}$$

Since the lenght of the path is smaller, the rope will not be stretched anymore.

*The angle between the force and radial vectors has a complex expression, however. We again suppose the bar is turned by an angle of $\theta$ (now counterclockwise):

Maths

We notice that the tangent of $\alpha$ is equal to $\frac{d-r\sin\theta}{r\cos\theta}$. Our angle is, however, the sum $\alpha+\theta$: $$x=\arctan{\left(\frac{d-r\sin\theta}{r\cos\theta}\right)}+\theta=\arctan{\left(\frac{d}{r\cos\theta}-\tan\theta\right)} + \theta$$

This is enough for a general solution.

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  • $\begingroup$ Yes the motor would have to be able to move back and forth in order to keep the tension in the pulley. $\endgroup$ – ja72 Jun 1 '15 at 14:51

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