# Conservation of Energy of a pendulum

1. Aug 14, 2011

### knowNothing23

A pendulum consists of a 2.0-kg bob attached to a light 3.0-m-long
string. While hanging at rest with the string vertical, the bob is struck a sharp
horizontal blow, giving it a horizontal velocity of 4.5 m/s. At the instant the string
makes an angle of 30° with the vertical, what is (a) the speed, (b) the gravitational
potential energy (relative to its value is at the lowest point), and (c) the tension in
the string? (d) What is the angle of the string with the vertical when the bob
reaches its greatest height?

My problem is with the answer of c.)
The book shows a different answer.
Thank you.

3. The attempt at a solution
I have taken the free body diagram of the pendulum and the string, when the string forms 30 degrees with the vertical.
The net force sum of the "y" component:
$\sumF$=Tcos(30)-mg=m(v^(2) )/ L , where "L" is the length of the string.

Solving for T:
T=(m/cos(30) )( v^(2)/L +g)

2. Aug 14, 2011

### PeterO

Provided you have used that equation correctly it should get the right answer - assuming you had part (a) correct.

In case it is the way you are using the formula poorly - evaluate this

T = [mg + (m.v^2)/L]/cos30

3. Aug 15, 2011

### knowNothing23

My answer in part a.) is equal to the one provided by the book. The answer is 3.5171 m/s.

The book's answer of problem c.) is 25N.

4. Aug 15, 2011

### PeterO

I think you [and me] were looking at this incorrectly.

T = (mv^2 / L) + mgcos(theta) not (mv^2 / L) + mg/cos(theta)

The mv^2 / L part we agree with, I shall ignore that.

The mg/cos(theta) parts comes from analysing a conical pendulum situation, when the acceleration is horizontal.
HOWEVER, this time the bob is merely swinging - and importantly slowing down.
The net force for this part of the analysis is tangential to the arc, not horizontal.

Draw a "free body diagram" for that situation and you will see what is happening.

I don't know how to attach diagrams or pictures to this page - so I have tried to describe it.

5. Aug 16, 2011

### knowNothing23

I'm not sure, what you mean. In a conic situation there is a centripetal acceleration as well as a tangential. Also, the equation I proposed is not the one you mentioned.
T=(m/cos(30) ) ( v^(2)/L) +g), the cosine of 30 degrees divides mv^(2)/L as well.

There are more than two ways to analyze the sum of vectors at the angle of 30 degrees and both ways should give the same result. We can either take the component of T or the component of Weight and sum them. In my approach I took the component of T and subtracted Weight from it.

6. Aug 16, 2011

### ehild

The direction of the centripetal force is the same as that of the tension in the string. It is not vertical.

The usual procedure is to write the force components parallel and perpendicular to the string. The parallel component of gravity is mgcos(30°). The centripetal force is equal to the resultant of the tension and the radial component of gravity:

Mv^2/L = T-mgcos(30°) Use v=3.5171 m/s.

ehild

7. Aug 17, 2011

### knowNothing23

Thank you, ehild.
That was my question. Why take the component of the weight, when one can do it for the tension.

8. Aug 18, 2011

### ehild

See attached figures. The one on the left shows the forces acting on the bob: tension and gravity. The resultant is shown by blue. It is not parallel with the string as the bob has both velocity and acceleration along the arc. The resultant force is equal to the centripetal force (along the string) + the tangential force, tangent to the arc T+G = Fcp+Ft (all vectors).Writing this equation in x and y components, you get:

T cos(30°)-mg=Fcp cos(30°)-Ft sin(30°)
-T sin(30°) = -Fcp sin(30°)-Ft cos(30°)

Solve. If you do it correctly, you get T=Fcp+mgcos(30°). A bit more complicated than the other one, with radial and tangential components, is not it?

ehild

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9. Aug 18, 2011

### ehild

The net force is neither horizontal nor tangential: it has a component along the arc and also the radial component: the centripetal force.

When writing a post, scroll down, click on "Manage Attachments"

ehild

10. Aug 20, 2011

Thank you!