Solve Circular Motion Help - Pendulum Bob Mass, Rod Length, Speed & Tension

[garyljc]

Circular motion help !

Hello ,

Here's my question :
A pendulum bob has mass 0.2kg. It is attached to one end of a light rod of length 2m. The rod is free to rotate in a vertical plane about an axis through theo ther end o. Given that the pendulum swings through 60degrees on either side of the vertical
a.) calculate the speed of the bob at the lowest point of its path
b.)the tension in the rod when the rod makes an angle of 30 degrees with the downward vertical

For part a.)
Using the work-energy principle :
v(square) = 2g
therefore v = 4.43m/s

I have no idea how to do part B
My first attempt for this :
Using work-energy principle (slotting r=2 , m =0.2 , theta=30,u=0) to find the velocity at that position , so that I can find the tension using
Tsin30 - 2g = m.v(square)/r

but i just can't get it =( , could anyone help me with tension equations ?

 

[Doc Al]

a.) calculate the speed of the bob at the lowest point of its path
...
For part a.)
Using the work-energy principle :
v(square) = 2g
therefore v = 4.43m/s

Use conservation of mechanical energy. Hint: At the top of the motion, the energy is purely potential.

 

[garyljc]

what do you mean conservation of mechanical energy ? well the question I'm doing doesn't describe to be a complete circle motion .
I got part A correct , but it's just part B that cause problems to me

 

[Doc Al]

Once you've found the speed, use Newton's 2nd law applied to the radial direction to find the tension. The acceleration is centripetal.

 

[Doc Al]

I have no idea how to do part B
My first attempt for this :
Using work-energy principle (slotting r=2 , m =0.2 , theta=30,u=0) to find the velocity at that position , so that I can find the tension using
Tsin30 - 2g = m.v(square)/r

I assume by "work-energy" principle you mean conservation of energy. In any case, that equation for radial forces is not quite right. Hint: What's the component of the weight in the radial direction?

 

[garyljc]

the current work-energy principle I'm learning for circular motion is
v(square) = u(square)-2gr(1-cos(theta))

but they said 30 degree to the downward vertical , it can be also left or right =( This will then affect my theta , and hence the calculations ?

If i take 30 to the left my equation will be something like :
Tcos(theta) - mg = m.v(square)/r ... is it correct ?

 

[Doc Al]

the current work-energy principle I'm learning for circular motion is
v(square) = u(square)-2gr(1-cos(theta))

This is just conservation of energy applied to this particular situation.

but they said 30 degree to the downward vertical , it can be also left or right =( This will then affect my theta , and hence the calculations ?

Does it really matter? Check and see.

If i take 30 to the left my equation will be something like :
Tcos(theta) - mg = m.v(square)/r ... is it correct ?

No, it's not. How did you arrive at this equation? (Realize that this equation has nothing to do with the earlier equation.) Analyze the forces acting on the bob. Find their components in the radial direction. Apply Newton's 2nd law.

 

[garyljc]

oh okay , my bad ... i'll think about it now , thanks anyway =-)

 

[garyljc]

hey doc , i still can't figure out , could you just help me out with this one ?

 

[Doc Al]

Just do what I asked in my last post:

Analyze the forces acting on the bob. Find their components in the radial direction. Apply Newton's 2nd law.​

 

[garyljc]

doc ,
i've tried to analyse the forces acting . this is what i got (though it doesn't fit the answer )

1. find v(sqaure) , by inserting (theta) as 30 , using the work energy principle
2. then solve the equation Tcos30-0.2g = m.v(square)/r

r=2m , m=0.2

could you see what's wrong with my equations ?

 

[Doc Al]

2. then solve the equation Tcos30-0.2g = m.v(square)/r

That equation is not right. Do this:
- List the forces that act on the bob (there are only two)
- Find the components of those forces in the radial direction (parallel to the rod)
- Set the sum of those force equal to ma, where a is the centripetal acceleration ($a_c = m v^2 / r$)

 

[garyljc]

Another question
does it matter if i find the compenent parallel to the rod or parallel to the weight ?

 

[Doc Al]

I'll answer that question with a question: Which direction does the centripetal acceleration point?