Minimum velocity at bottom of pendulum

[deep838]

Homework Statement

A pendulum is made of a rigid rod (mass m, length l) and a small bob of mass M attached at one end. The rod is pivoted on the other end. What should be the minimum speed of the bob at its lowest point so that the pendulum completes a full circle?

Homework Equations

anything that'll work

The Attempt at a Solution

I thought of finding the energy of the system when the bob is at the highest position. It came as E=(M+m/4)v²₁/2 where v₁ is the velocity of the bob at topmost position. Then to get the energy at the bottom position, and then equating the two, this is what I get:
(2M+m)gl + (M+m/4)v²₁/2 = (M+m/4)v²₂/2 , where v₂ is the velocity of the bob at the bottom position

But then what?

 

[Curious3141]

Homework Statement

A pendulum is made of a rigid rod (mass m, length l) and a small bob of mass M attached at one end. The rod is pivoted on the other end. What should be the minimum speed of the bob at its lowest point so that the pendulum completes a full circle?

Homework Equations

anything that'll work

The Attempt at a Solution

I thought of finding the energy of the system when the bob is at the highest position. It came as E=(M+m/4)v²₁/2 where v₁ is the velocity of the bob at topmost position. Then to get the energy at the bottom position, and then equating the two, this is what I get:
(2M+m)gl + (M+m/4)v²₁/2 = (M+m/4)v²₂/2 , where v₂ is the velocity of the bob at the bottom position

But then what?

At the lowest possible speed of the bob at the bottom of the swing that still allows full circle oscillation, the bob and the rod would have lost all kinetic energy at the top of the swing and therefore be at momentary rest. So there's no need to consider KE when bob is at the highest position - it's zero. All kinetic energy has been converted to gravitational PE at the top of the swing.

 

[deep838]

At the lowest possible speed of the bob at the bottom of the swing that still allows full circle oscillation, the bob and the rod would have lost all kinetic energy at the top of the swing and therefore be at momentary rest. So there's no need to consider KE when bob is at the highest position - it's zero. All kinetic energy has been converted to gravitational PE at the top of the swing.

So then E(top) = 2Mgl+mgl/2 and E(top)=E(bottom)=(M+m/4)v²/2

Solving v=√4gl and this is one of the options in the answer.

Thanks.

 

[Curious3141]

So then E(top) = 2Mgl+mgl/2 and E(top)=E(bottom)=(M+m/4)v²/2

Solving v=√4gl and this is one of the options in the answer.

Thanks.

Your PE equation looks wrong. Remember that the differences in height between the top and bottom are 2l for the bob M and l for the rod m (centre of mass of rod m has moved by height ½*l*2 = l). It's like going from the 6 o' clock to the 12 o' clock position.

Better recheck your working. I couldn't get the mass terms to cancel out.

 

[deep838]

:P right! But then doing 1/2 * (M+m/4)* v^2 = 2Mgl + mgl gives v=√[8(2M+m)/(4M+m)] that's not an option! am I going wrong somewhere else?

 

[Curious3141]

:P right! But then doing 1/2 * (M+m/4)* v^2 = 2Mgl + mgl gives v=√[8(2M+m)/(4M+m)] that's not an option! am I going wrong somewhere else?

I'm sorry, I think I was on the wrong track.

The conservation equation should be (bottom) rotational KE of rod + translational KE of bob = (top) gravitational PE of rod + gravitational PE of bob.

The rod's moment of inertia $\displaystyle I$ about its end (pivot) is given by $\displaystyle \frac{1}{3}ml^2$. Its rotational KE is then $\displaystyle \frac{1}{2}I\omega^2$, where $\displaystyle \omega = \frac{v}{L}$, where $\displaystyle v$ is the speed of the bob.

You still can't cancel out the masses, but I think this should be the right approach. The rod is not a point mass, and rotational kinetic energy cannot be ignored.

Also, can you please list the choices you have?

 

[deep838]

ok... will try that, meanwhile, these are my choices:

1. sqrt [ g*l*(12M+6m)/(3M+m) ]
2. sqrt [ 4*g*l ]
3. sqrt [ 5*g*l ]
4. sqrt [ g*l*(15M+6m)/(3M+m)

 

[Curious3141]

ok... will try that, meanwhile, these are my choices:

1. sqrt [ g*l*(12M+6m)/(3M+m) ]
2. sqrt [ 4*g*l ]
3. sqrt [ 5*g*l ]
4. sqrt [ g*l*(15M+6m)/(3M+m)

Yes, the answer is an option on that list. It's either 1 or 4 because I've told you we can't cancel out the masses, now you just have to determine which it is.

 

[deep838]

doing as you said,
1/2 v^2(M+m/6) = Mg*2l + mg*l ... is the equation which gives v= sqrt [ g*l*(12M+6m)/(3M+m/2) ] which is close to option 1, but with (3M + m/2) in the denominator instead of (3M+ m).

Is there still something wrong?

 

[Curious3141]

doing as you said,
1/2 v^2(M+m/6)

Shouldn't the rotational kinetic energy of the rod be $\displaystyle \frac{1}{2}I\omega^2 = (\frac{1}{2})(\frac{1}{3})ml^2 \omega^2 = \frac{1}{6}mv^2$?

So total KE of bob and rod = $\displaystyle \frac{1}{2}Mv^2 + \frac{1}{6}mv^2 = \frac{1}{6}(3M + m)v^2$.

 

[deep838]

Shouldn't the rotational kinetic energy of the rod be $\displaystyle \frac{1}{2}I\omega^2 = (\frac{1}{2})(\frac{1}{3})ml^2 \omega^2 = \frac{1}{6}mv^2$?

So total KE of bob and rod = $\displaystyle \frac{1}{2}Mv^2 + \frac{1}{6}mv^2 = \frac{1}{6}(3M + m)v^2$.

Yeah, that's what I did, but made a simple calculation error! Option 1 is coming to be correct. Thanks for your time and help.