Tension in a Pendulum Problem - 30kg, 4m cord, 4m/s @ bottom

[VinnyCee]

The ball has a mass of 30 kg and a speed of 4\,\frac{m}{s} at the instant it is at its lowest point, \theta\,=\,0. Determine the tension in the cord and the rate at which the ball's speed is decreasing at the instant \theta\,=\,20. Neglect the size of the ball. NOTE: The cord length is 4 m.

Here is what I have found so far, the acceleration:

\sum\,F_t\,=\,-W\,sin\,\theta\,=\,m\,a_t

-m\,g\,sin\,\theta\,=\,m\,a_t

a_t\,=\,-g\,sin\,\theta

a_t\,=\,-\left(-9.81\,\frac{m}{s^2}\right)\,sin\,(20)

a_t\,=\,3.36\,\frac{m}{s^2}

I cannot get the right answer for the tension in the cord though. Please help!

 

[Doc Al]

What did you get for the tension in the cord? (Show how you solved it.)

 

[VinnyCee]

\sum\,F_{\theta}\,=\,T\,-\,W\,=\,m\,a

T\,=\,m\,a\,+\,W

T\,=\,(30\,kg)\,\left(3.36\,\frac{m}{s^2}\right)+\,(30\,kg)\,\left(9.81\,\frac{m}{s^2}\right)

T\,=\,395\,N

The answer in the back of the book is T\,=\,361\,N though.

 

[Hootenanny]

HINT: Think centripetal force and energy.

 

[VinnyCee]

I thought that was what I was doing?

please help, I have no idea what it is that I am doing wrong.

 

[Hootenanny]

T\,=\,(30\,kg)\,\left({\color{red}3.36}\,\frac{m}{s^2}\right)+\ ,(30\,kg)\,\left(9.81\,\frac{m}{s^2}\right)

What you have there is the tangental acceleration, which is not the same an the centripetal acceleration. The centripetal acceleration is given by;

a_{c} = \frac{v^2}{r}

You can calculate the velocity using conservation of energy. You will need to use the velocity at the bottom of the swing and the change in height. Note, that at any point in the swing the resultant force must equal the centripetal force;

\sum\vec{F} = m\frac{v^2}{r} = T - mg\cos\theta

Can you go from here?

 

[VinnyCee]

Here is how I solved for the velocity at the instant:

s\,=\,\theta\,r

ds\,=\,d\theta\,r

-g\,sin\,\theta\,=\,a_t\,=\,v\,\frac{dv}{ds}\,=\,v\,\frac{dv}{d\theta\.r}

-g\,r\,sin\,\theta\,d\theta\,=\,v\,dv

-g\,r\,\int_{0}^{\theta}\,sin\,\theta\,d\theta\,=\,\int_0^v\,v\,dv

v^2\,=\,2\,g\,r\,\left(cos\,\theta\,-\,1\right)

v\,=\,\sqrt{2\,\left(-9.81\,\frac{m}{s^2}\right)\,(4\,m)\,\left[cos\,(20)\,-\,1\right]}\,=\,2.18\,\frac{m}{s}

Then I used the above result for this equation below:

T\,=\,(30\,kg)\frac{\left(2.18\,\frac{m}{s}\right)^2}{4\,m}\,+\,(30\,kg)\,\left(9.81\,\frac{m}{s^2}\right)\,\cos\,(20)

t\,=\,35.64\,+\,276.6\,=\,312.3\,N

This answer is still not correct though!

 

[Hootenanny]

Perhaps conservation of energy, would be a easier approach than calculus? I'll start;

\frac{1}{2}mv_{i}^{2} = \frac{1}{2}mv_{f}^{2} + mgh

v_{i}^{2} = v_{f}^{2} + 2gh

Now, using trig we find that h is given by;

h = 4 - 4\cos\theta

If you can't see this, try drawing a diagram where the pendulum is displaced from the equilibrium position. Thus, we obtain;

v_{i}^{2} = v_{f}^{2} + 2g( 4 - 4\cos\theta)

You need to solve for v_f. Can you go from here?

 

[VinnyCee]

What is the definition of h?

I used your v_f in the above T equation and it is still an incorrect answer. Is the T equation right?

 

[Hootenanny]

What is the definition of h?

I used your v_f in the above T equation and it is still an incorrect answer. Is the T equation right?

h is change in vertical height. Yes, your tension equation is correct. I think you have made an arithmatic error somewhere, as I obtained 361N using the above method which agrees with the answer given. I'll take the next step for you;

T = m\frac{v_{i}^{2} - 2g(r - r\cos\theta)}{r} + mg\cos\theta

Just sub your number directly in. Can you go from here?

 

[VinnyCee]

I found something about pendulums and the definition of h: http://en.wikipedia.org/wiki/Pendulum

T = (30 kg)\frac{\left(4 \frac{m}{s}\right)^{2} - 2\left(9.81 \frac{m}{s^2}\right)\left[(4 m) - (4 m)\cos (20)\right]}{4 m} + (30 kg)\left(9.81 \frac{m}{s^2}\right)\cos (20)

T = (30 kg) \frac{11.26 \frac{m^2}{s^2}}{4 m} + 276.6 N

T = 84.50 N + 276.6 N

T = 361 N

Thats cool! But how do I solve it using the other method that was mentioned?

 

[Hootenanny]

I found something about pendulums and the definition of h: http://en.wikipedia.org/wiki/Pendulum

T = (30 kg)\frac{\left(4 \frac{m}{s}\right)^{2} - 2\left(9.81 \frac{m}{s^2}\right)\left[(4 m) - (4 m)\cos (20)\right]}{4 m} + (30 kg)\left(9.81 \frac{m}{s^2}\right)\cos (20)

T = (30 kg) \frac{11.26 \frac{m^2}{s^2}}{4 m} + 276.6 N

T = 2055N?

I will say again check your calculations, you are using the correct the formula and are plugging in the correct numbers; you are simply making an arithmetic error. Try doing the calculation is stages.