Solving Pendulum Motion: Vo, Angle 48', Horizontal & Vertical Positions

[Inkyspider]

1. There is a pendulum of Length 2.1m. Its bob has speed Vo when the cord makes the angle 48' with the vertical.
a) what is the speed of bob in lowest posotioin if Vo=8m/s?
What is the least value Vo could have is pendulum is to swing down and then up to:
b) a horizontal position
c) a vertical position with cord remaining straight



2. I understand that this is a conservation of energy problem and K1 + U1 = K2 + U2 and i tried using this with K=1/2mv^2 and U=mgy but it didn't work.
Also, i know the answer to c involves something to do with tension, but I'm not sure how to use it.


Thanks for any help!

 

[Doc Al]

2. I understand that this is a conservation of energy problem and K1 + U1 = K2 + U2 and i tried using this with K=1/2mv^2 and U=mgy but it didn't work.

Show what you did.

Also, i know the answer to c involves something to do with tension, but I'm not sure how to use it.

Hint for c: Analyze the forces on the bob when it is in the vertical position and apply Newton's 2nd law to find the minimum speed required at that position to maintain a straight cord.

 

[kerimek]

a) To find the speed of the bob in its lowest position, we can use the conservation of energy principle. At the highest point, all of the energy is in the form of potential energy (U=mgh), and at the lowest point, all of the energy is in the form of kinetic energy (K=1/2mv^2). Therefore, we can set these two energies equal to each other and solve for v:

U = K
mgh = 1/2mv^2
gh = 1/2v^2
v = √(2gh)

Plugging in the given values, we get:

v = √(2*9.8*2.1) = 6.49 m/s

b) To swing down and then up to a horizontal position, the bob will need to have enough speed to overcome the gravitational force and reach a height of 0 at the highest point. Therefore, we can use the same equation as before, but set the potential energy at the highest point to 0:

U = K
mgh = 1/2mv^2
0 = 1/2v^2
v = 0

So, the least value for Vo in this case would be 0 m/s.

c) To swing down and then up to a vertical position with the cord remaining straight, the bob will need to have enough speed to reach a height equal to the length of the pendulum (2.1m) at the highest point. In addition, the tension in the cord will also play a role in the motion. The equation for this scenario can be written as:

U + T = K
mgh + T = 1/2mv^2

To solve for v, we need to know the tension in the cord (T). This can be found using the equation T = mgcosθ, where θ is the angle between the cord and the vertical. In this case, θ = 48', so cosθ = cos(48') = 0.7431.

Plugging in the values, we get:

mgh + T = 1/2mv^2
mgh + mgcosθ = 1/2mv^2
mg(h + cosθ) = 1/2mv^2
v = √(2g(h + cosθ))

Plugging in the given values, we get