# Rotational Vs Linear Acceleration

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1. Feb 15, 2016

### Eric V

Hi guys, I'm having a debate with a mechanical engineer friend of mine, and I was wondering if you could help me solve it. I'm not much of a physicist, but honestly I think he might have this one wrong, I just can't remember my old physics classes well enough to calculate and be sure.

The question is this: Does an object falling straight down strike with more force than an object which was held up at the end of a support and allowed to fall/swing through a pivot point?

Lets pretend we are using a golf ball. If we hold that golf ball three feet in the air and drop it, ignoring air resistance, how fast is it going when it hits the ground vs

If we had a pivot point 3 feet in the air, and attached a golf ball to the end of a 3 foot long zero mass support, rotated that support up around the circumference to the traditional pi radians position and allowed it to swing downward to the 1.5pi radians position, how fast would it be going?

Then the crux of the question is if you let the x axis of the circle, a line drawn through the points at 0 radians and pi radians, be ground level, and the the point at pi/2 radians be our starting position, is the ball going faster if you let it drop through the radius directly to the centerpoint of the circle, or if you let it swing down from pi/2 to pi radians using the radius of the circle as a pivot? And is the equation governing the velocity(or should I be more concerned about the acceleration?) uniform throughout every position on the circle?

Thank you for your help, I know it's not the most interesting question.

2. Feb 15, 2016

### Eric V

I was just reading the rules at the top of this forum- please give me a heads up if this question is too homework like. I'd like to have the opportunity to copy it and move it to the homework section.

3. Feb 15, 2016

### jbriggs444

As I understand it, you are comparing three scenarios:

Ball dropping three feet straight down
Ball following circular trajectory from "twelve o'clock" down to three or nine.
Ball following circular trajectory from three or nine o'clock down to six.

For the circular cases, the ball is constrained to the circular path by a massless rod which exerts no torque. It is like a bob on the end of a massless pendulum.

What does conservation of energy tell you about the final speed?

Can you determine what the acceleration will be at twelve o'clock, three o'clock and six o'clock? Are those accelerations identical?

4. Feb 15, 2016

### Eric V

I really can't. Like I was saying, I'm a computer science major. I haven't done any physics since '09. My thoughts are:

Perhaps the speed would be exactly the same?

Perhaps they would have the same vector in the y axis, but the ball at the end of a pendulum would have a vector pulling it crossways along the x axis, perhaps resulting in more acceleration as it approaches the 6 or 3/9 oclock?

I'm not sure which is the case, or if both of those are wrong.

5. Feb 15, 2016

### jbriggs444

In these forums, we try to help people work things out themselves rather than just providing answers. I was a computer science major myself and do IP networking for a living. It was '75 when I took my last physics course. Let's return to that conservation of energy argument

The rod/pendulum is massless. It has no potential energy and no kinetic energy.
The axle to which the rod/pendulum is mounted is motionless. Its potential energy is unchanging and it has no kinetic energy.
The only object that matters in an energy analysis is the ball/bob.

We can assume negligible air resistance and a friction-free axle. There is no outside interaction which might cause energy not to be conserved.

Regardless of which scenario we choose, the ball/bob drops by three feet. Therefore it loses the same amount of potential energy in each scenario. It follows that it must gain the same kinetic energy in each scenario. It starts at rest in all three scenarios. Will the final speeds in each of the three scenarios be the same or different?

6. Feb 15, 2016

### Eric V

It would have to be the same. Neither myself nor my friend were right. I was imagining the ball at the end of the pivot as having to accelerate faster around the curve in order to travel the same distance downward, and my friend felt that the pivot would slow the path of the ball and result in less force.

What equations governing motion would apply to this situation? I would like to do the calculations to demonstrate the answer.

7. Feb 15, 2016

### jbriggs444

The equations of motion for the circular case are most easily found by doing the analysis in terms of angular acceleration and torque.

The torque supplied by gravity is given by the force due to gravity multiplied by the current horizontal offset of the ball from the pivot point.

$\tau = mgr\ sin(\theta)$​

where $\tau$ is the torque, m is the mass of the object, g is the acceleration of gravity, r is the length of the pendulum and $\theta$ is the angle that the pendulum makes with the vertical.

The angular acceleration is given by the equation

$\tau = I\alpha$​

where $\tau$ is torque, I is the moment of inertia of the pendulum/bob assembly and $\alpha$ is the angular acceleration in radians per second2. This is the rotational analog for Newton's second law.

For a mass on the end of a massless rod, the moment of inertia is given by $mr^2$ and we can substitute that into the acceleration equation to give

$\tau = mr^2\alpha$​

Putting that together with the first equation, and dividing both sides by mr we have

$g\ sin(\theta) = r\alpha$​

That is a differential equation. In this form, it is beyond my ability to solve. But for small angles, one can make the approximation that $sin(\theta)$ is approximately $\theta$. Making this approximation and rewriting $\alpha$ as the second derivative of $\theta$ gives

$g\theta = r\frac{d^2 \theta}{{dt}^2}$​

or

$\theta - \frac{r}{g}\frac{d^2 \theta}{{dt}^2} = 0$​

That's a first (errr, second, actually) order linear homogeneous differential equation. Those are easy to solve.

Near the top of the circle (at the 12 o'clock position), the sign convention works out so that the solution to this is an exponential (actually the sum of a decaying exponential plus an increasing exponential in proportions that depend on the exact initial conditions) for $\theta$ measured from the top.

Near the bottom of the circle, the sign convention works out so that the solution to this is simple harmonic motion (the sum of sine and cosine terms in proportions that depend on the initial conditions) for $\theta$ measured from the bottom.

I may have muffed the equations along the way, but the characterization of the result should be correct.

Last edited: Feb 15, 2016
8. Feb 15, 2016

### Khashishi

Your friend is right if you include friction.

9. Feb 15, 2016

### Eric V

Thank you so much for your help on this. The process you did of going through all of the equations was above and beyond what I expected.

10. Feb 15, 2016

### Eric V

We were simplifying the situation as we have done here.