Pendulum Swinging Speeds: Solving for Initial Velocity and Angle

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The discussion revolves around calculating the speed of a pendulum bob at various positions based on its initial velocity and angle. For part (a), the speed at the lowest point was calculated as 8.35 m/s. In part (b), it was clarified that the minimum speed at the horizontal position could be zero, while for part (c), the minimum speed at the top of the swing must be sufficient to maintain tension in the string, leading to a derived formula. The conversation highlights the importance of understanding forces acting on the bob to solve for the initial velocity required to reach specific angles. Overall, the calculations emphasize energy conservation and the role of gravitational forces in pendulum motion.
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Homework Statement


A pendulum of length ##L=1.25m##.Its bob (which effectively has all the mass) has speed ##v_0## when the cord makes an angle ##θ_0=40.0^°## with the vertical.##(a)## What is the speed of the bob when it is in its low (est position If ##v_0=8.00m/s## ? What is the least value that ##v_0## can have If the pendulum is to swing down and then up ##(b)## to a horizontal position,and ##(c)## to a vertical position with the chord remaining straight ? ##(d)## Do the answers to parts ##(b)## and ##(c)## increase,decrease,or remain the same if ##θ_0## is increased by a few degrees ?

Homework Equations


##-ΔU=W##
##W=Δ(KE)##

The Attempt at a Solution


[/B]
I found the answer of (a) using -
##-ΔU=W##
##W=Δ(KE)##
it came ##v=8.35\frac m s##

I didnt understand (b) and part (c).so I stucked there...

Thanks
 

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Arman777 said:
I didnt understand (b)
Do the same calculation as you did for part a) to get an expression for the speed v at any given angle, θ, in terms of v0, θ0, L, g and θ.
 
##\sqrt {(v_0)^2+2gL(cosθ-cosθ_0)}=v##
 
Arman777 said:
##\sqrt {(v_0)^2+2gL(cosθ-cosθ_0)}=v##
Ok. So what is θ in b)?
 
90 degree
 
Arman777 said:
90 degree
So plug that in. What is the smallest value v0 can have?
 
Ok I found but for c what would be the angle ?
 
Arman777 said:
Ok I found but for c what would be the angle ?
θ0? Same as before.
 
haruspex said:
θ0? Same as before.

θ ?
 
  • #10
Arman777 said:
θ ?
You told me in post #5.
I have the feeling that there is something you are not understanding about the question, and I can't figure out what it is.
 
  • #11
I found the answer of (b),plotting θ=90
For (c) what is θ=?
 
  • #12
Arman777 said:
I found the answer of (b),plotting θ=90
For (c) what is θ=?
I'm very sorry, my eyes kept jumping to the wrong line, so I was reading b instead of c.
For c there is a complication. For b, the speed at the horizontal position could be 0. Could the speed be zero at the top of the loop?
 
  • #13
haruspex said:
I'm very sorry, my eyes kept jumping to the wrong line, so I was reading b instead of c.
For c there is a complication. For b, the speed at the horizontal position could be 0. Could the speed be zero at the top of the loop?

It cant
 
  • #14
Arman777 said:
It cant
Right. So what is the minimum speed at the top.
 
  • #15
##v=2\sqrt {gL}##
 
  • #16
Arman777 said:
##v=2\sqrt {gL}##
No. Consider this condition
Arman777 said:
with the cord remaining straight
What forces act on the bob at the top? What must the net force be?
 
  • #17
haruspex said:
No. Consider this condition

What forces act on the bob at the top? What must the net force be?

Yeah I made a mistake I know I noticed now
 
  • #18
v=6.72
 
  • #19
Arman777 said:
v=6.72
That's not right for the speed atthe top. How did you get that? Or is that your answer for v0?
 
  • #20
At the bottom total energy is 1/2mv^2 which here v is as we found ot 8.35 and at the top energy is 1/2mv^2+mg2L so I thought the should be equal
 
  • #21
Arman777 said:
At the bottom total energy is 1/2mv^2 which here v is as we found ot 8.35 and at the top energy is 1/2mv^2+mg2L so I thought the should be equal
For c, you are finding the initial velocity such that it will reach the top with a straight string. You do not know the KE earlier.
Forget energy for the moment and think about forces. What is the minimum speed at the top for the string to remain straight?
 
  • #22
T-mg=mv2/r
 
  • #23
Arman777 said:
T-mg=mv2/r
Right. What is the minimum value of T?
 
  • #24
I solved thanks :)
 

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