What is the speed of a ball on a pendulum at the lowest point?

Question

A heavy ball swings on a string in a circular arc of radius 2 m.

The two highest points are Q and Q' where the string is 29 degrees from the vertical. Point P is the lowest point where the string is vertically down.

acceleration due to gravity is 9.8 m/s^2.

What is the ball's speed at point P? Neglect air resistance and other frictional forces.

A heavy ball swings on a string in a circular arc of radius 2 m.

The two highest points are Q and Q' where the string is 29 degrees from the vertical. Point P is the lowest point where the string is vertically down.

acceleration due to gravity is 9.8 m/s^2.

What is the ball's speed at point P? Neglect air resistance and other frictional forces.

What is the magnitude of the ball's acceleration at point P?

Answer 1

mgh = 1/2 mv2
gravitational potential at top of swing = kinetic at bottom of swing
Find the vertical component that the ball is raised with the 29 degrees and the 2m radius, plug into equation, solve for v. Masses will cancel out. Since at the lowest point, all the gravitational potential is converted into kinetic energy.

Answer 2

Simple solution you know the ropes length and that at 29 degrees from the vertical the ball is at rest so V=0 Now using geometry you can find the vertical distance the ball falls

I would draw a diagram but I cant needless to say draw a picture and youll find the the answer to the vert distance traveled is
( 2 ) -( cos 29 degrees ) *( 2 ) = 0.2507 meters

Vfinal ^ 2 = Vinit ^ 2 + (2* a *x)
a= acceleration
x = distance
Vint=0
V final= what you are trying to find
Use this formula to find the answer for V final
V final = 2.217 m/s

Now do your own homework.... lazy @ss

Answer 3

the simplest thank you to to do it somewhat is to equate the flair capability of the pendulum to the kinetic capability on the backside element. The ball is raised via a distance equivalent to: 40 six(a million-cos(24))=3.98cm its ability capability is comparable to: m.g.h = 510*9.80 one*(3.ninety 8/a hundred)= 199.12 N-m its kinetic capability on the backside element of the trajectory is= 0.5(m.v^2) =0.5*510*v^2=199.12 v^2=199.12/(0.5*510)=0.seventy 8 v=sqrt(0.seventy 8)=0.88m/s The ball will swing to the comparable perspective to the different area = 24 levels.

Answer 4

Explanation: At Point Q and Q', the ball has maximum gravitational potential energy as they are at their highest points. At P, it is at its lowest point, and it is moving at the fastest speed at this point, so its kinetic energy is the max here while its gravitational potential energy is zero.

We need the height, h, of ball at point Q from a horizontal plane at the lowest point of the pendulum.

So, cos 29 = x/2
x = 1.749m
hence h = 2 - 1.749 = 0.251m

let g= 10N/kg
Mgh = 2.51xM

Since Gravitational Potential energy and kinetic energy are both max, they are equal.

2.51M = 0.5 Mv^2
so v = sqrt. 5.02
= 2.24m/s

Hope this helps=)

Answer 5

I don't know anything about this but would the length of the string make a difference?