Kinetic and Potential energy problem

[BrainMan]

Homework Statement

A simple pendulum is constructed by supporting a .375 kg steel sphere by a thin string 1.2 in length and of negligible mass. The pendulum is released form rest at a position where the string makes an angle of 35° with the vertical. What is the speed of the sphere as it reaches the lowest point of its arc.

Homework Equations

KE= 1/2mv^2
PE=mgy

The Attempt at a Solution

I attempted to find the potential energy when it is at the top of its path and then ,since it will have the same amount of energy at the bottom, find the velocity using the formula for KE (air friction I believe is not taken into account). I unfortunately don't know how to find the height of the sphere based on the angle from the vertical.

 

[dauto]

Draw a picture and use Geometry/Trigonometry

 

[BrainMan]

Do I have to use circle geometry? If so what is the relationship between the y distance, the radius, and the vertical angle?

 

[SammyS]

Do I have to use circle geometry? If so what is the relationship between the y distance, the radius, and the vertical angle?

Use trigonometry to come up with that.

 

[Regtic]

if angle B was 35° then $a$ would be (c)(cosB)

Hope that helps

 

[BrainMan]

What I tried to do now was to make a right triangle and do 1.2 cos theta to get the upper distance and then subtract that number from 1.2 to get x. I still got the wrong answer.

 

[BrainMan]

The correct answer is .323 m/s

 

[haruspex]

View attachment 69860

What I tried to do now was to make a right triangle and do 1.2 cos theta to get the upper distance and then subtract that number from 1.2 to get x. I still got the wrong answer.

That's the right method. Please post the details of your working.

 

[BrainMan]

1.2 cos 37 = .958
1.2 - .958 = .242
mgy = .889
1/2mv^2= .889
v^2 = 4.74
v = 2.18

 

[haruspex]

1.2 cos 37

It isn't 37.

 

[BrainMan]

I still am not getting the correct answer even though I am using 35°. It still comes out larger than the answer.

 

[haruspex]

I still am not getting the correct answer even though I am using 35°. It still comes out larger than the answer.

Strictly speaking, you are using the wrong length for the pendulum. You need to add the radius of the sphere. However, that involves looking up the density of steel to figure out the radius.
Similarly, not all the energy goes into linear motion. The sphere rotates as it swings. That also appears to involve finding the radius, but I think you'll find it cancels out in this case.
Try taking the rotation into account but continuing to use 1.2m for the length.

 

[BrainMan]

Strictly speaking, you are using the wrong length for the pendulum. You need to add the radius of the sphere. However, that involves looking up the density of steel to figure out the radius.
Similarly, not all the energy goes into linear motion. The sphere rotates as it swings. That also appears to involve finding the radius, but I think you'll find it cancels out in this case.
Try taking the rotation into account but continuing to use 1.2m for the length.

What do you mean by taking the rotation into account?

 

[haruspex]

What do you mean by taking the rotation into account?

I mean as in $\frac 12 I \omega^2$.
But let me correct an error in my previous post - I don't think the radius will cancel out. That happens in the case of a rolling sphere, but not here. So I would suggest: Let the radius of the sphere be r and obtain a purely symbolic expression for the answer. Don't plug any numbers in yet. Post what you get, maybe an outline of the working too. Let's get that right before putting in the numbers.

 

[BrainMan]

Can you clarify what that formula means? I am not aware of what the symbols represent.

 

[BrainMan]

Is that the formula for the KE?

 

[haruspex]

Is that the formula for the KE?

Yes, it's the formula for rotational KE. I is the moment of inertia about the point of rotation, ω is the angular speed.
I recommend treating the sphere (at its lowest point) as having horizontal linear motion speed v at the sphere's centre, and rotating about that centre.
Do you know the formula for moment of inertia of a uniform solid sphere about its centre?
Can you write out the relationship between v, ω, r and l?

 

[BrainMan]

No I do not

 

[haruspex]

No I do not

For the moment of inertia see http://hyperphysics.phy-astr.gsu.edu/hbase/isph.html.
What about the relationship between v, ω, r and l?

 

[BrainMan]

I unfortunately am having trouble understanding what your approach to the problem is.

 

[Regtic]

I'm somewhat confident the book's answer is wrong, but I'm not a physics expert by any means. The correct answer if you take into account the radius of the sphere with a steel density of 7900kg/m³ would be 2.18 m/s (ironically the same speed you calculated when you mistook the angle to be 37°) but if you ignore it, it should be 2.06 m/s. The method you took was correct.

Shouldn't have to take rotation into account here because we know that at the lowest point it will have the maximum kinetic energy and the work energy theorem is path independent since energy is being conserved.

 

[BrainMan]

OK I have figured out this problem the reason I got it wrong the first time with trig is because I forgot to convert inches to meters. It turns out you don't have to do any rotational KE or find the radius of the sphere. Thanks everyone for your help!

 

[Regtic]

OK I have figured out this problem the reason I got it wrong the first time with trig is because I forgot to convert inches to meters. It turns out you don't have to do any rotational KE or find the radius of the sphere. Thanks everyone for your help!

Man I was suspicious of that since the length you gave didn't have units while everything else did. I tried plugging in some different values but it didn't occur to me to try non SI units since I'm used to Canadian physics problems.

 

[BrainMan]

I figured out why my first attempt didn't work. I didn't convert from inches to meters. It turns out that you don't have to use the rotational KE or the radius of the sphere.

 

[haruspex]

I'm somewhat confident the book's answer is wrong, but I'm not a physics expert by any means. The correct answer if you take into account the radius of the sphere with a steel density of 7900kg/m³ would be 2.18 m/s

I get 2.08 m/s. The ball is 2.2 cm in radius, yes?
But I agree, the book answer is way off. I can't see what simple error leads to it.

Shouldn't have to take rotation into account here because we know that at the lowest point it will have the maximum kinetic energy and the work energy theorem is path independent since energy is being conserved.

I don't understand your reasoning. Even at the lowest point, the ball is rotating as well as moving horizontally. However, it turns out that the energy taken up by that is only 0.01% of the total, so it can be safely ignored. It would be a very different story if the radius were similar in magnitude to the length of the string.

 

[Regtic]

I get 2.08 m/s. The ball is 2.2 cm in radius, yes?
But I agree, the book answer is way off. I can't see what simple error leads to it.
I don't understand your reasoning. Even at the lowest point, the ball is rotating as well as moving horizontally. However, it turns out that the energy taken up by that is only 0.01% of the total, so it can be safely ignored. It would be a very different story if the radius were similar in magnitude to the length of the string.

woops you're right, hit square root instead of cube root I think. As I said, I'm no expert. When I did these kinds of problems in my mechanics class we never took rotation into account. If we took rotation into account, wouldn't the energy be the exact same anyways?
$\frac {1}{2} Iω^2 = \frac {1}{2}(mr^2) \frac {v^2}{r^2} = \frac {1}{2} mv^2$ no?

 

[haruspex]

If we took rotation into account, wouldn't the energy be the exact same anyways?
$\frac {1}{2} Iω^2 = \frac {1}{2}(mr^2) \frac {v^2}{r^2} = \frac {1}{2} mv^2$ no?

No, you're forgetting the length of the string. $\omega = \frac v{l+r}$. The total KE is $\frac {1}{2} Iω^2 + \frac {1}{2}mv^2 = \frac {1}{2} \frac 25 m r^2 ω^2 + \frac {1}{2}mv^2 = \frac {m}{2} v^2(\frac 25 \left(\frac r{l+r}\right)^2 + 1)$

 

[Regtic]

No, you're forgetting the length of the string. $\omega = \frac v{l+r}$. The total KE is $\frac {1}{2} Iω^2 + \frac {1}{2}mv^2 = \frac {1}{2} \frac 25 m r^2 ω^2 + \frac {1}{2}mv^2 = \frac {m}{2} v^2(\frac 25 \left(\frac r{l+r}\right)^2 + 1)$

Lolwut. I've done like maybe 30 problems involving torque or angular velocity if we discount my waves course. Is there a good link that you know of about this relationship? I have too many questions to reasonably ask you to answer like: Isn't the radius the distance from the center of mass to the point that we're rotating about so r in this case accounts for the length of the string? Where did the 2/5 come from? What's the $(\frac {r}{l+r})^2$? Where did the extra v^2 come from on the angular energy in the last equation. Aren't we counting the kinetic energy twice?

Don't bother trying to answer all of that, I'll look it all up after finals are over.

 

[haruspex]

Lolwut. I've done like maybe 30 problems involving torque or angular velocity if we discount my waves course. Is there a good link that you know of about this relationship? I have too many questions to reasonably ask you to answer like: Isn't the radius the distance from the center of mass to the point that we're rotating about so r in this case accounts for the length of the string? Where did the 2/5 come from? What's the $(\frac {r}{l+r})^2$? Where did the extra v^2 come from on the angular energy in the last equation. Aren't we counting the kinetic energy twice?

Don't bother trying to answer all of that, I'll look it all up after finals are over.

You can treat the motion of the sphere as either a rotation about the support point of the string or as a linear motion of the sphere's centre plus a rotation about that centre.
The angular speed is the same either way. The linear speed of the sphere's centre is v = (l+r)ω.
The moment of inertia of a uniform solid sphere mass M radius r about its centre is 2Mr²/5. Standard formula.
If we treat it as a linear motion plus a rotation then we must add the KEs:
$\frac 12 \frac 25 M r^2 \omega^2 + \frac 12 M v^2 = \frac 12 M (\frac 25 \frac {r^2 }{(l+r)^2} v^2 + v^2) = \frac 12 M v^2(\frac 25 \frac {r^2 }{(l+r)^2} +1)$
Treating it as a rotation about the support point, we need the moment of inertia about that point. For this we use the parallel axis theorem, which tells us we have to add the mass of the sphere times the square of the displacement, giving 2Mr²/5 + M(l+r)².
$\frac 12 (\frac 25 M r^2 + M(l+r)^2)\omega^2$ gives the same result.

 

[Regtic]

You can treat the motion of the sphere as either a rotation about the support point of the string or as a linear motion of the sphere's centre plus a rotation about that centre.
The angular speed is the same either way. The linear speed of the sphere's centre is v = (l+r)ω.
The moment of inertia of a uniform solid sphere mass M radius r about its centre is 2Mr²/5. Standard formula.
If we treat it as a linear motion plus a rotation then we must add the KEs:
$\frac 12 \frac 25 M r^2 \omega^2 + \frac 12 M v^2 = \frac 12 M (\frac 25 \frac {r^2 }{(l+r)^2} v^2 + v^2) = \frac 12 M v^2(\frac 25 \frac {r^2 }{(l+r)^2} +1)$
Treating it as a rotation about the support point, we need the moment of inertia about that point. For this we use the parallel axis theorem, which tells us we have to add the mass of the sphere times the square of the displacement, giving 2Mr²/5 + M(l+r)².
$\frac 12 (\frac 25 M r^2 + M(l+r)^2)\omega^2$ gives the same result.

Oooh ok. Ya that makes sense, tried to understand moment of innertia off a quick wiki read. mr² isn't the moment of innertia. I just read the link you posted up top about innertia. I assumed it was the radius of the circle it's rotating around. I haven't ever even heard of the parallel axis theorem, I'll give it a look after finals but the other part made perfect sense. Thank you.