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Here's my problem diagram:

Pendulum

I am working on a hobby project in which a motor is required to rotate a pulley that is connected to a pendulum as shown in the image. The maximum torque, Horsepower and RPM ratings of the motor are to be determined so that it can move the pendulum object of mass $m$ for a distance of $L$ and height $h$. The rivet on the pulley is at a distance $d$ from the equilibrium position of the pendulum.

This is my attempt so far.

Force required by the pulley, $$F = m * a$$ Torque, $$T = F * r $$

Where I am stuck - how to calculate a??

Once I get the Torque, I know how to calculate the Horsepower.

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closed as off-topic by ACuriousMind, Bill N, AccidentalFourierTransform, John Rennie, CuriousOne Apr 27 '16 at 6:36

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  • 2
    $\begingroup$ A pendulum moves all by itself. That's the whole idea: supply it with potential or kinetic energy once and it does a periodic motion. How fast or slow you do that doesn't matter, hence it doesn't make sense to ask "how much power is needed". $\endgroup$ – CuriousOne Apr 25 '16 at 5:50
  • $\begingroup$ As drawn the pendulum bob will never reach the intended position. The length of the pendulum will be an important parameter in any analysis. As a start forget about the acceleration of the bob and treat it as a statics problem. As the rivet rotates the line of action of the tension in the string of length $d$ comes closer to the centre of the pulley. The static torque is the tension in that string times the perpendicular distance between the line of action of the tension and the centre of the pulley. $\endgroup$ – Farcher Apr 25 '16 at 7:29
  • $\begingroup$ This problem is much more complicated than you think. As a static problem, the torque required increases with h, and will be high if d is not much greater than L. As a dynamic problem, the torque will vary with time and depend on the frequency of rotation of the pulley and the natural frequency of swing of the pendulum. If the two are equal, no torque is required at all. If the connecting links are strings rather than rigid rods this adds further complications because strings can pull but not push. $\endgroup$ – sammy gerbil Apr 25 '16 at 10:18
  • $\begingroup$ My advice is that if your project is to build this device rather than solve the equations of motion mathematically, it is easier to find suitable values for hp, m, L, etc by trial and error, eg starting with a small prototype. This will also show you what kinds of complex motion can result. $\endgroup$ – sammy gerbil Apr 25 '16 at 10:22
  • $\begingroup$ Do you know the mass moment of inertia of the pulley, motor and any gear-train attached to it? The dynamics of the system might depend on this a lot. $\endgroup$ – ja72 Apr 26 '16 at 19:51
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Hint: first find the angular velocity of the disc by conservation of energy,

$\frac{1}{2}I\omega^2 = mgh$ , $\omega$ = angular velocity of the pulley, I = moment of inertia of the pulley.

It is mentioned that the pendulum cover a horizontal distance L in t seconds. And we also know that $$v_{linear-velocity} = R\omega$$

So time t can be found by:

$$\dfrac{L}{R\omega}$$ where R is radius of the pulley.

Now you can find angular acceleration as you know both w and t for the particular instance. Now torque of the pulley can be found out because:

torque required for the pulley = (moment of inertia of the pulley)* ( angular acceleration)

Now when the motor swings the pendulum to its extreme position, only then the torque is required and remember the above method will give average torque required by the pendulum to move from its mean to the extreme position. Once the pendulum is moved to its extreme position, it will require no torque to move to its mean position. Hence to continue the motion in equilibrium state only that much torque would be required which would just overcome the negative torque provided by the motor.

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  • $\begingroup$ This answer would be much better with math formatting. Enclose expressions in dollar signs$...$ for inline or $$...$$ for paragraph style. Then use \frac{a}{b} for fractions like $\frac{a}{b}$. $\endgroup$ – ja72 Apr 26 '16 at 17:44

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